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Worked solutions

IB Computer Science SL · Paper 1

Computer Science, Standard Level

Timezone

TZ1TZ2

Session: May 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 May 2023 TZ1 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

Role and core functionality of a web browser

Networks

2 marks

HTTP request handling: Translates URLs and sends HTTP/HTTPS GET requests to retrieve remote web resources.
Rendering engine: Interprets HTML markup, applies CSS rules, and executes client-side scripts to render formatted web pages.

Browser HTTPS GET Server HTML/CSS Display

Mentor tip

Explain both fetching remote web documents and rendering HTML on screen.

Question 2

1 mark

Role of the Memory Address Register (MAR) in CPU architecture

Computer organisation

1 mark

The Memory Address Register (MAR) holds the physical memory address in primary storage (RAM) that is currently being accessed for a read or write operation.

MAR Address Bus RAM

Mentor tip

MAR holds the memory address; MDR holds the data contents.

Question 3

2 marks

Universal character encoding via Unicode

Computer organisation

2 marks

Global linguistic coverage: Supports character glyphs across virtually all human languages, scripts, and symbol sets, overcoming the 128-character limit of 7-bit ASCII.
Cross-platform text consistency: Ensures text renders identically across diverse operating systems and international environments without data corruption.

Unicode: > 140, 000 characters across UTF-8, UTF-16

Mentor tip

State that Unicode encodes characters from international alphabets and symbols.

Question 4

4 marks

Primary storage characteristics and high-speed cache memory

Computer organisation

(a)

1 mark

Volatility: RAM is volatile memory that loses all stored data when the power supply is interrupted.

Mentor tip

State volatility or direct byte-level addressability by the CPU.

(b)

3 marks

High-speed buffer: Cache is ultra-fast SRAM situated directly on or near the CPU die, operating much faster than main DRAM.
Locality of reference: Holds copies of frequently used instructions and data words.
Latency mitigation: Reduces average memory access latency by satisfying requests directly from cache, avoiding slow bus cycles to primary RAM.

CPU ⟷ L1/L2/L3 Cache ⟷ RAM

Mentor tip

Explain that cache stores frequently accessed instructions to reduce memory access delays.

Question 5

4 marks

Truth table construction for compound logic expression

Computer organisation

4 marks

Evaluate expression: $Z = A nand (B nor C)$ .
Evaluate sub-term $B nor C$ : Outputs 1 only when $B = 0$ and $C = 0$ ; otherwise 0.
Evaluate NAND with $A$ : Outputs 0 only when $A = 1$ and $(B nor C) = 1$ ; otherwise 1.
Evaluate across all 8 combinations:
Row $A = 0, B = 0, C = 0 ⟹ 0 nand 1 = 1$ .
Row $A = 0, B = 0, C = 1 ⟹ 0 nand 0 = 1$ .
Row $A = 0, B = 1, C = 0 ⟹ 0 nand 0 = 1$ .
Row $A = 0, B = 1, C = 1 ⟹ 0 nand 0 = 1$ .
Row $A = 1, B = 0, C = 0 ⟹ 1 nand 1 = 0$ .
Row $A = 1, B = 0, C = 1 ⟹ 1 nand 0 = 1$ .
Row $A = 1, B = 1, C = 0 ⟹ 1 nand 0 = 1$ .
Row $A = 1, B = 1, C = 1 ⟹ 1 nand 0 = 1$ .

A 00001111 B 00110011 C 01010101 B nor C 10001000 A nand (B nor C) 11110111

Mentor tip

Only one input state outputs 0: A=1, B=0, C=0.

Question 6

2 marks

Abstract data structure properties of collections

Abstract data structures

2 marks

Dynamic sizing: A collection is an elastic container whose capacity grows and contracts dynamically as elements are added or removed.
Sequential access: Elements are accessed sequentially through iterator primitives (`getNext()`, `hasNext()`, `resetNext()`) without fixed numeric indices.

Mentor tip

Mention dynamic size and sequential access via iterator methods.

Question 7

2 marks

Layer identification in the OSI 7-layer networking model

Networks

2 marks

Identify two distinct layers from the standard OSI 7-layer model: Physical Layer and Data Link Layer (or Network Layer, Transport Layer, Session Layer, Presentation Layer, Application Layer).

OSI Layers: Application, Presentation, Session, Transport, Network, Data Link, Physical

Mentor tip

Name any two of the seven OSI reference model layers.

Question 8

6 marks

Reverse-index array algorithm: trace table and purpose deduction

Computational thinking, problem-solving and programming

(a)

4 marks

Given: $DATA [0..5] = [7, 21, 5, 19, 6, 2]$ . Initialize $K = 5, A = 1, B = 0$ .
Trace loop iterations step by step while $K \geq 0$ :
Iteration 1 ( $K = 5$ ): $DATA [5] = 2$ . $2 mod 2 = 0 < 1$ (even) $⟹ A = 1 \times 2 = 2$ . $K = 4$ .
Iteration 2 ( $K = 4$ ): $DATA [4] = 6$ . $6 mod 2 = 0 < 1$ (even) $⟹ A = 2 \times 6 = 12$ . $K = 3$ .
Iteration 3 ( $K = 3$ ): $DATA [3] = 19$ . $19 mod 2 = 1$ (odd) $⟹ B = 0 + 1 = 1$ . $K = 2$ .
Iteration 4 ( $K = 2$ ): $DATA [2] = 5$ . $5 mod 2 = 1$ (odd) $⟹ B = 1 + 1 = 2$ . $K = 1$ .
Iteration 5 ( $K = 1$ ): $DATA [1] = 21$ . $21 mod 2 = 1$ (odd) $⟹ B = 2 + 1 = 3$ . $K = 0$ .
Iteration 6 ( $K = 0$ ): $DATA [0] = 7$ . $7 mod 2 = 1$ (odd) $⟹ B = 3 + 1 = 4$ . $K = - 1$ .
Loop terminates ( $K = - 1 < 0$ ). Final outputs: $A = 12$ and $B = 4$ .

K 543210 - 1 DATA [K] 26195217 - A 21212121212 - B 001234 - Output 12, 4

Mentor tip

Track K decreasing from 5 down to 0, multiplying evens into A and tallying odds into B.

(b)

2 marks

The algorithm calculates and outputs the mathematical product of all even integers in the array (stored in $A$ ), and counts and outputs the total quantity of odd integers in the array (stored in $B$ ).

A = x \in DATA, x even \prod x = 2 \times 6 = 12, B = x \in DATA, x odd \sum 1 = 4

Mentor tip

Clearly describe the two outputs: product of even elements and count of odd elements.

Question 9

2 marks

Conceptual definition of computational abstraction

Computational thinking, problem-solving and programming

2 marks

Hiding extraneous details: Abstraction is the problem-solving technique of filtering out low-level, non-essential implementation details to focus on core high-level functional concepts.
Complexity management: Allows developers to construct modular software components using interfaces without having to manage internal underlying mechanics.

Mentor tip

Define abstraction as focusing on essential concepts while filtering out complex low-level implementation details.

Questions 10–12 · 45 marks

Question 10

15 marks

Legacy system migration, requirements engineering, changeover, and training

System fundamentals

(a)

2 marks

User resistance and anxiety: Users accustomed to existing legacy routines may resist learning new software workflows.
Role redefinition: Changes in business logic or automation may shift daily tasks, requiring job reassignments or revised permissions.

Mentor tip

Address user resistance to change and the need to adapt to revised workflow roles.

(b)

2 marks

A legacy system is an outdated or obsolete hardware/software platform that remains in active operational service because it performs essential business functions and is risky or costly to replace.

Mentor tip

Note both obsolescence and ongoing mission-critical reliance.

(c)

1 mark

Direct stakeholder interviews / Questionnaires and surveys / Direct workplace observation.

Mentor tip

Name one standard requirement-gathering technique.

(d)

3 marks

Data format and schema mismatch: Old records may store dates, names, or numerical codes in formats unsupported by the modern relational database.
Risk of data loss or corruption: Incompatible fields can result in truncated strings or dropped records during automated batch extraction unless validation transforms are applied.

Mentor tip

Highlight schema incompatibility and risk of lost or corrupted data.

(e)

3 marks

Zero operational disruption: Running both systems concurrently ensures that if the new system encounters unexpected defects, operations continue seamlessly on the proven legacy system without business downtime.

Mentor tip

Focus on risk mitigation and seamless fallback.

(f)(i)

1 mark

Self-paced interactive online tutorials / In-person instructor-led workshops.

Mentor tip

Choose self-paced online training or instructor-led workshops.

(f)(ii)

3 marks

Advantages: Learners progress at their individual speed, can repeat lessons as needed, and study around their working schedule.
Disadvantages: Lacks immediate personal feedback from an expert instructor, and requires strong individual motivation.

Mentor tip

Balance schedule flexibility against lack of immediate expert support.

Question 11

15 marks

VPN architecture, speed factors, data compression, and remote telecommuting

Networks
System fundamentals

(a)

2 marks

Tunneling protocols (e.g. IPsec, OpenVPN) to encapsulate private network packets within public network traffic.
Cryptographic data encryption (e.g. AES) to guarantee confidentiality of data in transit.

Mentor tip

Name encryption and tunneling protocols.

(b)

2 marks

Channel bandwidth: The maximum bit-rate capacity supported by the physical or wireless medium.
Network traffic congestion: Heavy concurrent traffic causing router queue delays and packet drops.

Mentor tip

State bandwidth and network traffic congestion.

(c)

3 marks

Reduced file size: Compression removes redundancy, minimizing the total number of bytes sent.
Faster transfer times: Transmitting fewer bytes takes less time, conserving bandwidth and accelerating downloads.

Transmission Time = \frac{Compressed Size}{Bandwidth}

Mentor tip

Explain smaller byte payload leading to faster transmission and bandwidth savings.

(d)

2 marks

Plaintext scrambling: Transforms readable data into ciphertext using a mathematical algorithm and secret key.
Interception protection: Intercepted packets remain completely unreadable without the corresponding decryption key.

Plaintext Encrypt Ciphertext

Mentor tip

Detail scrambling into ciphertext using cryptographic keys.

(e)

2 marks

Rule-based filtering: Inspects incoming/outgoing packet headers against predefined security rules, blocking unauthorized connections while allowing legitimate traffic.

Mentor tip

State rule-based packet inspection and blocking unauthorized access.

(f)

4 marks

Positive impacts: Greater working flexibility, elimination of stressful daily commutes, and improved ability to manage domestic responsibilities.
Negative impacts: Social isolation from colleagues, blurring of work-life boundaries leading to overwork, and reduced spontaneous collaborative brainstorming.

Mentor tip

Discuss both positive aspects (flexibility, zero commute) and negative aspects (isolation, blurred boundaries).

Question 12

15 marks

Campus room billing management: searching algorithms and descending bubble sort

Abstract data structures
Computational thinking, problem-solving and programming

(a)

2 marks

Contiguous storage: Elements are stored in consecutive primary memory locations.
Direct index access: Any element can be accessed in constant time $O (1)$ using its numeric index.

Address (A [i]) = Base + i \times Size

Mentor tip

Mention contiguous memory allocation and constant-time $O (1)$ indexing.

(b)

2 marks

Linear search (sequential search) and Binary search (half-interval divide-and-conquer search).

Linear Search: O (n) vs Binary Search: O (lo g n)

Mentor tip

Name Linear search and Binary search.

(c)

5 marks

Input search target $X$ .
Initialize tracking flag `FOUND = false` and index $I = 0$ .
Iterate through the array while $I < 10$ and not `FOUND`:
If $ROOMNUMS [I] == X$ , set `FOUND = true`.
Increment index $I$ .
Output appropriate confirmation message based on `FOUND` flag state.

sub-program check (ROOMNUMS) input X FOUND = false I = 0 loop while I < 10 and not FOUND if ROOMNUMS [I] == X then FOUND = true end if I = I + 1 end loop if FOUND then output ("The student in room ", X, " has paid the bill") else output ("The student in room ", X, " has not paid the bill yet") end if end sub-program

Mentor tip

Use a sequential scan with early termination once the room number is found.

(d)

6 marks

Descending sort requirement: Larger elements must bubble toward index 0, with smaller elements moving to the end.
Comparison swap condition: If $ROOMNUMS [I] < ROOMNUMS [I + 1]$ , swap the two elements so the smaller element shifts rightward.
Maintain a boolean swap flag until no adjacent inversions occur throughout a complete pass.

SWAPPED = true loop while SWAPPED SWAPPED = false loop I from 0 to 8 if ROOMNUMS [I] < ROOMNUMS [I + 1] then TEMP = ROOMNUMS [I] ROOMNUMS [I] = ROOMNUMS [I + 1] ROOMNUMS [I + 1] = TEMP SWAPPED = true end if end loop end loop

Mentor tip

Notice the comparison operator is '<' to achieve descending order (largest to smallest).

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 programming21 marks · 30%
Q8, Q9, Q12
System fundamentals19 marks · 27%
Q10, Q11
Networks15 marks · 21%
Q1, Q7, Q11
Computer organisation11 marks · 16%
Q2, Q3, Q4, Q5
Abstract data structures4 marks · 6%
Q6, Q12

Mentor insights

Traps to watch for

01
Truth table evaluation with NOR and NAND · Q5
Remember that B NOR C is 1 only when B=0 and C=0. Then A NAND 1 produces 0 only when A=1. All other 7 combinations evaluate to 1.
02
Decreasing loop counter with modulo checking · Q8
Notice K begins at 5 and decrements to 0. Modulo by 2 produces 0 for even values and 1 for odd values. Ensure even numbers multiply into A and odd numbers increment B.
03
Descending bubble sort swap condition · Q12
Standard bubble sort uses '>' for ascending order. For descending order (largest first), the swap condition must be inverted to ROOMNUMS[I] < ROOMNUMS[I+1].

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

Timezone

TZ1TZ2

Session: May 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 May 2023 TZ1 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

Role and core functionality of a web browser

Networks

2 marks

HTTP request handling: Translates URLs and sends HTTP/HTTPS GET requests to retrieve remote web resources.
Rendering engine: Interprets HTML markup, applies CSS rules, and executes client-side scripts to render formatted web pages.

Browser HTTPS GET Server HTML/CSS Display

Mentor tip

Explain both fetching remote web documents and rendering HTML on screen.

Question 2

1 mark

Role of the Memory Address Register (MAR) in CPU architecture

Computer organisation

1 mark

The Memory Address Register (MAR) holds the physical memory address in primary storage (RAM) that is currently being accessed for a read or write operation.

MAR Address Bus RAM

Mentor tip

MAR holds the memory address; MDR holds the data contents.

Question 3

2 marks

Universal character encoding via Unicode

Computer organisation

2 marks

Global linguistic coverage: Supports character glyphs across virtually all human languages, scripts, and symbol sets, overcoming the 128-character limit of 7-bit ASCII.
Cross-platform text consistency: Ensures text renders identically across diverse operating systems and international environments without data corruption.

Unicode: > 140, 000 characters across UTF-8, UTF-16

Mentor tip

State that Unicode encodes characters from international alphabets and symbols.

Question 4

4 marks

Primary storage characteristics and high-speed cache memory

Computer organisation

(a)

1 mark

Volatility: RAM is volatile memory that loses all stored data when the power supply is interrupted.

Mentor tip

State volatility or direct byte-level addressability by the CPU.

(b)

3 marks

High-speed buffer: Cache is ultra-fast SRAM situated directly on or near the CPU die, operating much faster than main DRAM.
Locality of reference: Holds copies of frequently used instructions and data words.
Latency mitigation: Reduces average memory access latency by satisfying requests directly from cache, avoiding slow bus cycles to primary RAM.

CPU ⟷ L1/L2/L3 Cache ⟷ RAM

Mentor tip

Explain that cache stores frequently accessed instructions to reduce memory access delays.

Question 5

4 marks

Truth table construction for compound logic expression

Computer organisation

4 marks

Evaluate expression: $Z = A nand (B nor C)$ .
Evaluate sub-term $B nor C$ : Outputs 1 only when $B = 0$ and $C = 0$ ; otherwise 0.
Evaluate NAND with $A$ : Outputs 0 only when $A = 1$ and $(B nor C) = 1$ ; otherwise 1.
Evaluate across all 8 combinations:
Row $A = 0, B = 0, C = 0 ⟹ 0 nand 1 = 1$ .
Row $A = 0, B = 0, C = 1 ⟹ 0 nand 0 = 1$ .
Row $A = 0, B = 1, C = 0 ⟹ 0 nand 0 = 1$ .
Row $A = 0, B = 1, C = 1 ⟹ 0 nand 0 = 1$ .
Row $A = 1, B = 0, C = 0 ⟹ 1 nand 1 = 0$ .
Row $A = 1, B = 0, C = 1 ⟹ 1 nand 0 = 1$ .
Row $A = 1, B = 1, C = 0 ⟹ 1 nand 0 = 1$ .
Row $A = 1, B = 1, C = 1 ⟹ 1 nand 0 = 1$ .

A 00001111 B 00110011 C 01010101 B nor C 10001000 A nand (B nor C) 11110111

Mentor tip

Only one input state outputs 0: A=1, B=0, C=0.

Question 6

2 marks

Abstract data structure properties of collections

Abstract data structures

2 marks

Dynamic sizing: A collection is an elastic container whose capacity grows and contracts dynamically as elements are added or removed.
Sequential access: Elements are accessed sequentially through iterator primitives (`getNext()`, `hasNext()`, `resetNext()`) without fixed numeric indices.

Mentor tip

Mention dynamic size and sequential access via iterator methods.

Question 7

2 marks

Layer identification in the OSI 7-layer networking model

Networks

2 marks

Identify two distinct layers from the standard OSI 7-layer model: Physical Layer and Data Link Layer (or Network Layer, Transport Layer, Session Layer, Presentation Layer, Application Layer).

OSI Layers: Application, Presentation, Session, Transport, Network, Data Link, Physical

Mentor tip

Name any two of the seven OSI reference model layers.

Question 8

6 marks

Reverse-index array algorithm: trace table and purpose deduction

Computational thinking, problem-solving and programming

(a)

4 marks

Given: $DATA [0..5] = [7, 21, 5, 19, 6, 2]$ . Initialize $K = 5, A = 1, B = 0$ .
Trace loop iterations step by step while $K \geq 0$ :
Iteration 1 ( $K = 5$ ): $DATA [5] = 2$ . $2 mod 2 = 0 < 1$ (even) $⟹ A = 1 \times 2 = 2$ . $K = 4$ .
Iteration 2 ( $K = 4$ ): $DATA [4] = 6$ . $6 mod 2 = 0 < 1$ (even) $⟹ A = 2 \times 6 = 12$ . $K = 3$ .
Iteration 3 ( $K = 3$ ): $DATA [3] = 19$ . $19 mod 2 = 1$ (odd) $⟹ B = 0 + 1 = 1$ . $K = 2$ .
Iteration 4 ( $K = 2$ ): $DATA [2] = 5$ . $5 mod 2 = 1$ (odd) $⟹ B = 1 + 1 = 2$ . $K = 1$ .
Iteration 5 ( $K = 1$ ): $DATA [1] = 21$ . $21 mod 2 = 1$ (odd) $⟹ B = 2 + 1 = 3$ . $K = 0$ .
Iteration 6 ( $K = 0$ ): $DATA [0] = 7$ . $7 mod 2 = 1$ (odd) $⟹ B = 3 + 1 = 4$ . $K = - 1$ .
Loop terminates ( $K = - 1 < 0$ ). Final outputs: $A = 12$ and $B = 4$ .

K 543210 - 1 DATA [K] 26195217 - A 21212121212 - B 001234 - Output 12, 4

Mentor tip

Track K decreasing from 5 down to 0, multiplying evens into A and tallying odds into B.

(b)

2 marks

A = x \in DATA, x even \prod x = 2 \times 6 = 12, B = x \in DATA, x odd \sum 1 = 4

Mentor tip

Clearly describe the two outputs: product of even elements and count of odd elements.

Question 9

2 marks

Conceptual definition of computational abstraction

Computational thinking, problem-solving and programming

2 marks

Hiding extraneous details: Abstraction is the problem-solving technique of filtering out low-level, non-essential implementation details to focus on core high-level functional concepts.
Complexity management: Allows developers to construct modular software components using interfaces without having to manage internal underlying mechanics.

Mentor tip

Define abstraction as focusing on essential concepts while filtering out complex low-level implementation details.

Questions 10–12 · 45 marks

Question 10

15 marks

Legacy system migration, requirements engineering, changeover, and training

System fundamentals

(a)

2 marks

User resistance and anxiety: Users accustomed to existing legacy routines may resist learning new software workflows.
Role redefinition: Changes in business logic or automation may shift daily tasks, requiring job reassignments or revised permissions.

Mentor tip

Address user resistance to change and the need to adapt to revised workflow roles.

(b)

2 marks

A legacy system is an outdated or obsolete hardware/software platform that remains in active operational service because it performs essential business functions and is risky or costly to replace.

Mentor tip

Note both obsolescence and ongoing mission-critical reliance.

(c)

1 mark

Direct stakeholder interviews / Questionnaires and surveys / Direct workplace observation.

Mentor tip

Name one standard requirement-gathering technique.

(d)

3 marks

Data format and schema mismatch: Old records may store dates, names, or numerical codes in formats unsupported by the modern relational database.
Risk of data loss or corruption: Incompatible fields can result in truncated strings or dropped records during automated batch extraction unless validation transforms are applied.

Mentor tip

Highlight schema incompatibility and risk of lost or corrupted data.

(e)

3 marks

Mentor tip

Focus on risk mitigation and seamless fallback.

(f)(i)

1 mark

Self-paced interactive online tutorials / In-person instructor-led workshops.

Mentor tip

Choose self-paced online training or instructor-led workshops.

(f)(ii)

3 marks

Advantages: Learners progress at their individual speed, can repeat lessons as needed, and study around their working schedule.
Disadvantages: Lacks immediate personal feedback from an expert instructor, and requires strong individual motivation.

Mentor tip

Balance schedule flexibility against lack of immediate expert support.

Question 11

15 marks

VPN architecture, speed factors, data compression, and remote telecommuting

Networks
System fundamentals

(a)

2 marks

Tunneling protocols (e.g. IPsec, OpenVPN) to encapsulate private network packets within public network traffic.
Cryptographic data encryption (e.g. AES) to guarantee confidentiality of data in transit.

Mentor tip

Name encryption and tunneling protocols.

(b)

2 marks

Channel bandwidth: The maximum bit-rate capacity supported by the physical or wireless medium.
Network traffic congestion: Heavy concurrent traffic causing router queue delays and packet drops.

Mentor tip

State bandwidth and network traffic congestion.

(c)

3 marks

Reduced file size: Compression removes redundancy, minimizing the total number of bytes sent.
Faster transfer times: Transmitting fewer bytes takes less time, conserving bandwidth and accelerating downloads.

Transmission Time = \frac{Compressed Size}{Bandwidth}

Mentor tip

Explain smaller byte payload leading to faster transmission and bandwidth savings.

(d)

2 marks

Plaintext scrambling: Transforms readable data into ciphertext using a mathematical algorithm and secret key.
Interception protection: Intercepted packets remain completely unreadable without the corresponding decryption key.

Plaintext Encrypt Ciphertext

Mentor tip

Detail scrambling into ciphertext using cryptographic keys.

(e)

2 marks

Rule-based filtering: Inspects incoming/outgoing packet headers against predefined security rules, blocking unauthorized connections while allowing legitimate traffic.

Mentor tip

State rule-based packet inspection and blocking unauthorized access.

(f)

4 marks

Positive impacts: Greater working flexibility, elimination of stressful daily commutes, and improved ability to manage domestic responsibilities.
Negative impacts: Social isolation from colleagues, blurring of work-life boundaries leading to overwork, and reduced spontaneous collaborative brainstorming.

Mentor tip

Discuss both positive aspects (flexibility, zero commute) and negative aspects (isolation, blurred boundaries).

Question 12

15 marks

Campus room billing management: searching algorithms and descending bubble sort

Abstract data structures
Computational thinking, problem-solving and programming

(a)

2 marks

Contiguous storage: Elements are stored in consecutive primary memory locations.
Direct index access: Any element can be accessed in constant time $O (1)$ using its numeric index.

Address (A [i]) = Base + i \times Size

Mentor tip

Mention contiguous memory allocation and constant-time $O (1)$ indexing.

(b)

2 marks

Linear search (sequential search) and Binary search (half-interval divide-and-conquer search).

Linear Search: O (n) vs Binary Search: O (lo g n)

Mentor tip

Name Linear search and Binary search.

(c)

5 marks

Input search target $X$ .
Initialize tracking flag `FOUND = false` and index $I = 0$ .
Iterate through the array while $I < 10$ and not `FOUND`:
If $ROOMNUMS [I] == X$ , set `FOUND = true`.
Increment index $I$ .
Output appropriate confirmation message based on `FOUND` flag state.

sub-program check (ROOMNUMS) input X FOUND = false I = 0 loop while I < 10 and not FOUND if ROOMNUMS [I] == X then FOUND = true end if I = I + 1 end loop if FOUND then output ("The student in room ", X, " has paid the bill") else output ("The student in room ", X, " has not paid the bill yet") end if end sub-program

Mentor tip

Use a sequential scan with early termination once the room number is found.

(d)

6 marks

Descending sort requirement: Larger elements must bubble toward index 0, with smaller elements moving to the end.
Comparison swap condition: If $ROOMNUMS [I] < ROOMNUMS [I + 1]$ , swap the two elements so the smaller element shifts rightward.
Maintain a boolean swap flag until no adjacent inversions occur throughout a complete pass.

SWAPPED = true loop while SWAPPED SWAPPED = false loop I from 0 to 8 if ROOMNUMS [I] < ROOMNUMS [I + 1] then TEMP = ROOMNUMS [I] ROOMNUMS [I] = ROOMNUMS [I + 1] ROOMNUMS [I + 1] = TEMP SWAPPED = true end if end loop end loop

Mentor tip

Notice the comparison operator is '<' to achieve descending order (largest to smallest).

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 programming21 marks · 30%
Q8, Q9, Q12
System fundamentals19 marks · 27%
Q10, Q11
Networks15 marks · 21%
Q1, Q7, Q11
Computer organisation11 marks · 16%
Q2, Q3, Q4, Q5
Abstract data structures4 marks · 6%
Q6, Q12

Mentor insights

Traps to watch for

01
Truth table evaluation with NOR and NAND · Q5
Remember that B NOR C is 1 only when B=0 and C=0. Then A NAND 1 produces 0 only when A=1. All other 7 combinations evaluate to 1.
02
Decreasing loop counter with modulo checking · Q8
Notice K begins at 5 and decrements to 0. Modulo by 2 produces 0 for even values and 1 for odd values. Ensure even numbers multiply into A and odd numbers increment B.
03
Descending bubble sort swap condition · Q12
Standard bubble sort uses '>' for ascending order. For descending order (largest first), the swap condition must be inverted to ROOMNUMS[I] < ROOMNUMS[I+1].

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