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VCE General Mathematics — Networks and decision mathematics Questions & Answers

Q38

2024

VCAA

Paper 1

1 mark

Q38

1 mark

A connected graph has six vertices and six edges.

How many of the following four statements must always be true?

the graph has no vertices of odd degree
the graph contains a Eulerian trail
the graph contains a Hamiltonian path
the sum of the degrees of the vertices is 12

A

1

B

2

C

3

D

4

Reveal Answer

A

1

Correct Answer

Only one statement is always true: the sum of the degrees is 12. By the Handshaking Lemma, the sum of degrees is always twice the number of edges ( $2 \times 6 = 12$ ). The other statements can be disproven with counterexamples.

B

2

Only one statement is always true. A graph consisting of a triangle with three separate edges attached to its vertices has degrees of 3, 3, 3, 1, 1, and 1, which disproves the first three statements.

C

3

Only one statement is always true. A connected graph with 6 vertices and 6 edges does not necessarily have a Eulerian trail or a Hamiltonian path, nor does it require all even degrees.

D

4

It is impossible for all four statements to always be true. Only the statement regarding the sum of the degrees being 12 is a universal property of any graph with 6 edges.

Q2

2024

QCAA

Paper 1

1 mark

Q2

1 mark

In a graph, an open walk with repeated vertices and no repeated edges is called a

A

bridge.

B

loop.

C

path.

D

trail.

Reveal Answer

A

bridge.

A bridge is a specific type of edge whose removal increases the number of connected components in a graph, not a type of walk.

B

loop.

A loop is an edge that connects a vertex to itself, rather than a sequence of vertices and edges known as a walk.

C

path.

A path is a walk where all vertices must be distinct. Since the question specifies that vertices are repeated, it cannot be a path.

D

trail.

Correct Answer

A trail is defined as a walk in which no edges are repeated, but vertices are allowed to be repeated.

Q5

2024

QCAA

Paper 2

6 marks

Q5

6 marks

A flying doctor coordinator allocates a plane from each of three airbases, A, B and C, to fly to one of three sites, P, Q and R, to provide medical care. Distances (km) are shown in the table.

	P ( $28^\circ$ S $136^\circ$ E)	Q	R ( $20^\circ$ S $147^\circ$ E)
A ( $20^\circ$ S $136^\circ$ E)	$x$	600	$y$
B	445	485	340
C	980	1170	770

Determine the optimal allocation for each plane and the minimum total distance flown.

Reveal Answer

Calculate $x$ (distance from A to P):
Angular distance $= 28^\circ - 20^\circ = 8^\circ$
$D = 111.2 \times 8^\circ \approx 890$ km

Calculate $y$ (distance from A to R):
Angular distance $= 147^\circ - 136^\circ = 11^\circ$
$D = 111.2 \times \cos 20^\circ \times 11^\circ \approx 1149$ km

Row reduction:
$\begin{bmatrix} 290 & 0 & 549 \\ 105 & 145 & 0 \\ 210 & 400 & 0 \end{bmatrix}$

Column reduction:
$\begin{bmatrix} 185 & 0 & 549 \\ 0 & 145 & 0 \\ 105 & 400 & 0 \end{bmatrix}$

Number of lines needed to cover all zeros = number of tasks ( $3 = 3$ ), so allocate planes.
For minimum distance, the plane allocation is airbase A to site Q, airbase B to site P and airbase C to site R.

Minimum total distance flown $= 600 + 445 + 770$
$= 1815$ km

Marking Criteria

Descriptor	Marks
correctly calculates distance x in kilometres	1
correctly calculates distance y in kilometres	1
reduces each row	1
reduces each column	1
identifies optimal allocation for each plane	1
determines minimum total distance flown	1

Q7

2024

QCAA

Paper 1

1 mark

Q7

1 mark

The table shows information for a project with four activities.

Activity	Duration (min)	Prerequisite	Earliest starting time	Latest starting time
W	1	—	0	4
X	2	—	0	0
Y	3	X	2	2
Z	4	W, Y	5	5

What is the float time for activity W, in minutes?

A

0

B

1

C

4

D

5

Reveal Answer

A

0

A float of 0 indicates a critical activity where the earliest and latest start times are identical. For activity W, these times differ.

B

1

This value represents the duration of activity W (1 minute), not the available float time.

C

4

Correct Answer

Float time is calculated as the difference between the Latest Starting Time (LST) and the Earliest Starting Time (EST). For activity W, $4 - 0 = 4$ minutes.

D

5

This value corresponds to the start times for activity Z, rather than the calculated float for activity W.

Q5

2025

QCAA

Paper 1

1 mark

Q5

1 mark

Identify the number of faces for a planar graph that has 5 edges and 3 vertices.

A

8

B

6

C

4

D

2

Reveal Answer

A

8

Incorrect. This incorrectly adds the number of vertices and edges ( $3 + 5 = 8$ ) instead of applying Euler's formula.

B

6

Incorrect. This is a miscalculation; applying Euler's formula ( $V - E + F = 2$ ) with $V=3$ and $E=5$ does not result in 6 faces.

C

4

Correct Answer

Correct. By Euler's formula for planar graphs ( $V - E + F = 2$ ), substituting $V = 3$ and $E = 5$ gives $3 - 5 + F = 2$ , which means there are $F = 4$ faces.

D

2

Incorrect. This is the constant value in Euler's formula ( $V - E + F = 2$ ), not the calculated number of faces.

Q20

2022

QCAA

Paper 1

4 marks

Q20

4 marks

The table summarises the distances in kilometres (km) between three flower stores and three delivery locations: A, B and C.

Use the Hungarian algorithm to determine the minimum total distance needed to deliver flowers to all locations if each store delivers flowers to only one location.

	A	B	C
Store 1	19	17	24
Store 2	15	14	22
Store 3	23	16	40

Reveal Answer

Row reduction
Store 1: 19 17 24 (-17) -> 2 0 7
Store 2: 15 14 22 (-14) -> 1 0 8
Store 3: 23 16 40 (-16) -> 7 0 24

$\begin{bmatrix} 2 & 0 & 7 \\ 1 & 0 & 8 \\ 7 & 0 & 24 \end{bmatrix}$

Column reduction (-1 0 -7)
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 6 & 0 & 17 \end{bmatrix}$

Therefore for minimum distance, Store 1 must deliver to C, Store 2 must deliver to A and Store 3 must deliver to B.

Minimum total distance = 24 + 15 + 16 = 55 km

Marking Criteria

Descriptor	Marks
correctly reduces each row	1
correctly reduces each column	1
identifies which store delivers to which location	1
determines minimum total distance	1

Q40

2025

VCAA

Paper 1

1 mark

Q40

1 mark

The precedence table below shows the 12 activities required to complete a project. The duration in days and immediate predecessors are shown.

Activity	Duration	Immediate predecessors
$A$	4	–
$B$	6	$A$
$C$	8	$A$
$D$	3	$A$
$E$	9	$B$
$F$	6	$C$
$G$	7	$B, D, F$
$H$	12	$C$
$I$	6	$G, H$
$J$	4	$E, I$
$K$	3	$G, H$
$L$	9	$J$

The project is to be completed in minimum time.

The float time, in days, of Activity $B$ is

A

4

B

6

C

8

D

12

Reveal Answer

A

4

This is the earliest start time (EST) of Activity $B$ , not its float time.

B

6

This is the duration of Activity $B$ , not its float time.

C

8

Correct Answer

The earliest start time for Activity $B$ is 4 and its latest finish time is 18. The float time is calculated as Latest Finish Time - Earliest Start Time - Duration, which is $18 - 4 - 6 = 8$ days.

D

12

This is the latest start time (LST) of Activity $B$ , not its float time.

Q8

2021

QCAA

Paper 1

1 mark

Q8

1 mark

A basketball competition has six teams that have completed three rounds of competition as shown.

	Bears	Eagles	Lions	Meerkats	Tigers	Wombats
Bears	—	—	✓	—	✓	✓
Eagles	—	—	✓	✓	—	✓
Lions	✓	✓	—	✓	—	—
Meerkats	—	✓	✓	—	✓	—
Tigers	✓	—	—	✓	—	✓
Wombats	✓	✓	—	—	✓	—

The graph to represent this information has

A

6 edges.

B

9 edges.

C

15 edges.

D

18 edges.

Reveal Answer

A

6 edges.

This number corresponds to the number of vertices (teams) in the graph, not the number of edges (games played).

B

9 edges.

Correct Answer

The table contains 18 checkmarks total. Since each game involves two teams and is recorded for both, the number of edges is half the total count: $18 \div 2 = 9$ .

C

15 edges.

This would be the number of edges if every team played every other team (a complete graph $K_6$ ), calculated as $\frac{6 \times 5}{2} = 15$ .

D

18 edges.

This is the total number of checkmarks in the table. Because the graph is undirected (Team A vs Team B is the same game as Team B vs Team A), you must divide this sum by 2.

Q7

2022

QCAA

Paper 1

1 mark

Q7

1 mark

This matrix was obtained after applying the Hungarian algorithm to determine the optimal allocation of three people, Elandra (E), Farid (F) and Grace (G), to three tasks: legal (L), monitoring (M) and verification (V).

\begin{matrix} & L & M & V \\ E & 0 & 0 & 7 \\ F & 0 & 3 & 8 \\ G & 1 & 0 & 0 \end{matrix}

The optimal allocation is

A

E to V, F to M and G to L.

B

E to V, F to L and G to M.

C

E to M, F to L and G to V.

D

E to M, F to V and G to L.

Reveal Answer

A

E to V, F to M and G to L.

This allocation is invalid because it assigns tasks based on non-zero entries in the matrix (e.g., E to V is 7), whereas the optimal solution must use only zero entries.

B

E to V, F to L and G to M.

Although F and G are assigned to zeros, this option assigns E to V, which has a value of 7 in the reduced matrix, indicating it is not part of the optimal solution.

C

E to M, F to L and G to V.

Correct Answer

Row F has only one zero at column L, forcing F to L. This leaves E with only one available zero at M, which in turn forces G to V, resulting in an allocation using only zeros.

D

E to M, F to V and G to L.

This allocation is suboptimal because it uses non-zero entries for F to V (value 8) and G to L (value 1).

Q8

2023

QCAA

Paper 1

1 mark

Q8

1 mark

Activities P and Q are the critical activities for a project.

Activity	Duration	Prerequisite activity
P	3	–
Q	6	P

What are the earliest starting time (EST) and latest starting time (LST) for Activity Q?

A

3 | 3

B

3 | 6

C

6 | 6

D

6 | 9

Reveal Answer

A

3 | 3

Correct Answer

Activity Q depends on P (duration 3), so its Earliest Start Time (EST) is 3. Since Q is a critical activity, it has zero float, meaning its Latest Start Time (LST) must equal its EST ( $3$ ).

B

3 | 6

The EST is correct, but the LST is wrong. Because Q is a critical activity, there is no slack, so the Latest Start Time cannot be later than the Earliest Start Time.

C

6 | 6

This incorrectly identifies the EST as 6. Since P starts at 0 and takes 3 units of time, Q can begin at time 3, not 6.

D

6 | 9

Both values are incorrect. The EST is determined by the completion of P (time 3), and since Q is critical, the LST must also be 3.

Q5

2020

QCAA

Paper 2

5 marks

Q5

5 marks

A company has three tasks to allocate to three contractors. Each of the contractors has a quote recorded for each task, shown in the table. The quotes are in thousands of dollars (＄'000s).

Contractor	Task 1	Task 2	Task 3
A	3	3	1
B	4	7	2
C	4	4	1

Use a matrix method to determine the minimum cost if each contractor is allocated one task.

Reveal Answer

Matrix form
$\begin{pmatrix} 3 & 3 & 1 \\ 4 & 7 & 2 \\ 4 & 4 & 1 \end{pmatrix}$

row reduction: $R_1 - 1, R_2 - 2, R_3 - 1$
$\begin{pmatrix} 2 & 2 & 0 \\ 2 & 5 & 0 \\ 3 & 3 & 0 \end{pmatrix}$

only need 1 line to cover all the 0s,
$\therefore$ column reduction: $C_1 - 2, C_2 - 2$
$\begin{pmatrix} 0 & 0 & 0 \\ 0 & 3 & 0 \\ 1 & 1 & 0 \end{pmatrix}$

need 3 lines to cover all the 0s, $\therefore$ bipartite graph:

contractor task cost
A 2 3
B 1 4
C 3 1
Total 8
$\therefore$ Minimum cost is ＄8000.

Marking Criteria

Descriptor	Marks
Correctly reduces each row	1
Correctly reduces each column	1
Allocates each task to one contractor	1
Determines minimum cost	1
Shows logical organisation, communicating key steps	1

Q3

2023

QCAA

Paper 1

1 mark

Q3

1 mark

The duration, in minutes, of all activities in a project are shown.

Activity	P	Q	R	S	T	U	V
Duration	38	42	32	34	16	14	26

The critical path for the project is PRSV.
What is the earliest completion time for the project if it starts at 11:00 am?

A

12:30 pm

B

1:10 pm

C

1:30 pm

D

2:10 pm

Reveal Answer

A

12:30 pm

This time implies a project duration of 90 minutes. However, the sum of the critical path activities (PRSV) is $38 + 32 + 34 + 26 = 130$ minutes.

B

1:10 pm

Correct Answer

The earliest completion time is determined by the total duration of the critical path PRSV. Summing the durations gives $38 + 32 + 34 + 26 = 130$ minutes (2 hours and 10 minutes). Adding this to the 11:00 am start time results in 1:10 pm.

C

1:30 pm

This time implies a duration of 150 minutes. The correct duration based on the critical path is 130 minutes.

D

2:10 pm

This time implies a duration of 190 minutes (3 hours and 10 minutes), which is one hour longer than the actual critical path duration of 130 minutes.

Q34

2023

VCAA

Paper 1

1 mark

Q34

1 mark

A bipartite graph is typically used to display which one of the following?

A

the allocation of tasks on a construction site

B

the path used to visit five different construction sites

C

the total distance travelled between two construction sites

D

the critical path of activities to be completed in a construction project

E

the minimum length of cable required to connect six construction sites

Reveal Answer

A

the allocation of tasks on a construction site

Correct Answer

Bipartite graphs are ideal for modeling assignment problems, such as allocating a set of workers to a set of tasks, where edges connect the two distinct groups.

B

the path used to visit five different construction sites

Finding a path to visit multiple sites is a routing problem (like the Travelling Salesperson Problem), which is typically modeled using a complete or general graph.

C

the total distance travelled between two construction sites

Determining the distance between two sites is a shortest path problem, which is best modeled using a weighted graph rather than a bipartite graph.

D

the critical path of activities to be completed in a construction project

Critical path analysis for project activities is modeled using a directed acyclic graph (DAG) or an activity network, as it requires showing dependencies and sequence.

E

the minimum length of cable required to connect six construction sites

Finding the minimum cable to connect multiple sites is a minimum spanning tree problem, which is modeled using a weighted graph.

Q1

2023

QCAA

Paper 2

5 marks

Q1

5 marks

A triathlon relay has three sections: swim (S), cycle (C) and run (R). The matrix shows the average number of minutes for three athletes, Jane (J), Knox (K) and Levi (L), to complete each section.

	S	C	R
J	40	56	66
K	36	60	72
L	25	48	78

Use the Hungarian algorithm to predict the minimum total relay time if assigning each athlete to completing one section.

Reveal Answer

$\begin{matrix} & \text{S} & \text{C} & \text{R} \\ \text{J} & 40 & 56 & 66 \\ \text{K} & 36 & 60 & 72 \\ \text{L} & 25 & 48 & 78 \end{matrix}$

column reduction $-25 \quad -48 \quad -66$

row reduction
$\begin{bmatrix} 15 & 8 & 0 \\ 11 & 12 & 6 \\ 0 & 0 & 12 \end{bmatrix} \begin{matrix} -0 \\ -6 \\ -0 \end{matrix}$

$\begin{bmatrix} 15 & 8 & 0 \\ 5 & 6 & 0 \\ 0 & 0 & 12 \end{bmatrix}$

Number of lines needed to cover all zeroes < number of tasks
$2 < 3$ , so continue algorithm steps.
Smallest uncovered number is 5. Subtract 5 from all uncovered numbers and add 5 to number covered twice.

$\begin{bmatrix} 10 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 17 \end{bmatrix}$

Number of lines needed to cover all zeroes = number of tasks
$3 = 3$ , so assign tasks.
To minimise the total relay time, assign Jane to run, Knox to swim and Levi to cycle.
Predicted minimum total relay time
$= 66 + 36 + 48$
$= 150 \text{ min}$
$= 2 \text{ h } 30 \text{ min}$

Marking Criteria

Descriptor	Marks
correctly reduces each column	1
reduces each row	1
continues algorithm steps until number of lines needed to cover all zeroes equals number of tasks	1
assigns each athlete to complete one section	1
predicts minimum total relay time including units	1

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