Networks — Worked Solutions (HSC Maths Standard 2)
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Worked examples for HSC Maths Standard 2 networks. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Standard 2 — Networks. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Shortest path
Question
A delivery network connects depot $A$ to warehouse $E$. The roads (in km) are:
$A\text{–}B = 4$, $A\text{–}C = 3$, $B\text{–}D = 5$, $C\text{–}B = 1$, $C\text{–}D = 6$, $D\text{–}E = 2$, $C\text{–}E = 9$.
Find the length of the shortest path from $A$ to $E$, and state the route.
Solution
Work out the shortest distance to each vertex from $A$, building up.
- To $C$: directly $A\text{–}C = 3$.
- To $B$: either $A\text{–}B = 4$, or $A\text{–}C\text{–}B = 3 + 1 = 4$. Shortest $= 4$.
- To $D$: via $B$, $4 + 5 = 9$; via $C$, $3 + 6 = 9$. Shortest $= 9$.
- To $E$: via $D$, $9 + 2 = 11$; direct $C\text{–}E$, $3 + 9 = 12$. Shortest $= 11$.
Shortest path is 11 km, route $A\text{–}C\text{–}D\text{–}E$ (or $A\text{–}B\text{–}D\text{–}E$, also 11). Always compare every way into a vertex before locking in its label.
The trick with shortest path is to find the shortest distance to each vertex in turn, always reusing the best distances you've already found.
Start at $A$. The nearest reachable vertex is $C$ at $3$ km. From there, $B$ can be reached as $A\text{–}C\text{–}B = 3 + 1 = 4$, which ties with the direct $A\text{–}B = 4$, so $B$ is $4$ km away.
Next, $D$: through $B$ it's $4 + 5 = 9$, and through $C$ it's $3 + 6 = 9$ — both give $9$ km. Finally $E$: via $D$ it's $9 + 2 = 11$, but the direct $C\text{–}E$ route is $3 + 9 = 12$, which is longer. So the shortest is 11 km along $A\text{–}C\text{–}D\text{–}E$. We compare routes into each vertex because a longer first step can still lead to a shorter overall path.
Label shortest distance to each vertex:
- $C$: $3$
- $B$: $\min(4,\ 3+1) = 4$
- $D$: $\min(4+5,\ 3+6) = 9$
- $E$: $\min(9+2,\ 3+9) = 11$
Shortest path $= 11$ km, route $A\text{–}C\text{–}D\text{–}E$.
Where the marks go
- 1 mark: Correct shortest distances to intermediate vertices $B$, $C$
- 1 mark: Correct shortest distance to $D$ (9 km)
- 1 mark: Correct shortest path of 11 km with a valid route
Key idea
Find the shortest distance to each vertex in turn, comparing every route into it; the answer reuses the best distances already established.
Example 2 — Minimum spanning tree
Question
A council wants to lay fibre cable connecting five parks $P$, $Q$, $R$, $S$ and $T$. The possible connections (in hundreds of metres) are:
$P\text{–}Q = 7$, $P\text{–}R = 5$, $Q\text{–}R = 9$, $Q\text{–}S = 8$, $R\text{–}S = 6$, $R\text{–}T = 4$, $S\text{–}T = 3$.
(a) Find the minimum spanning tree and state its total length.
(b) Explain why a minimum spanning tree connecting 5 parks always uses exactly 4 connections.
Solution
(a) Build the tree by adding the cheapest edge that doesn't form a cycle (Kruskal's method).
- $S\text{–}T = 3$ ✓ (connects $S$, $T$)
- $R\text{–}T = 4$ ✓ (adds $R$)
- $P\text{–}R = 5$ ✓ (adds $P$)
- $R\text{–}S = 6$ ✗ — $R$ and $S$ already linked, forms a cycle, skip
- $P\text{–}Q = 7$ ✓ (adds $Q$ — all 5 connected)
Total $= 3 + 4 + 5 + 7 = 19$ (i.e. 1900 m).
(b) A spanning tree on $n$ vertices always has $n - 1$ edges, so $5 - 1 = 4$. Reject any edge that closes a cycle — that's where marks go.
(a) A minimum spanning tree links every vertex as cheaply as possible with no loops. The neat way is to keep choosing the smallest available edge, skipping any that would create a cycle.
Sorted cheapest first: $S\text{–}T = 3$ (take it), $R\text{–}T = 4$ (take it, brings in $R$), $P\text{–}R = 5$ (take it, brings in $P$), then $R\text{–}S = 6$ — but $R$ and $S$ are already joined, so this would make a loop; skip it. Next is $P\text{–}Q = 7$, which finally connects $Q$. Now all five parks are linked.
Adding up: $3 + 4 + 5 + 7 = 19$, so 1900 m of cable.
(b) Think of it as connecting parks one at a time: the first edge joins 2 parks, and each new edge after that adds exactly one new park without forming a loop. To go from 1 connected park to 5 you add 4 edges — in general $n - 1$.
(a) Kruskal — add cheapest non-cycle edges:
- $S\text{–}T = 3$ ✓
- $R\text{–}T = 4$ ✓
- $P\text{–}R = 5$ ✓
- $R\text{–}S = 6$ ✗ (cycle)
- $P\text{–}Q = 7$ ✓ (all connected)
Total $= 3 + 4 + 5 + 7 = 19 = 1900$ m.
(b) A spanning tree on $n$ vertices has $n - 1$ edges; $5 - 1 = 4$.
Where the marks go
- 1 mark: Selects edges in increasing weight, avoiding cycles
- 1 mark: Correct set of MST edges (ST, RT, PR, PQ)
- 1 mark: Correct total length of 19 (1900 m)
- 1 mark: Explains a spanning tree on $n$ vertices has $n - 1$ edges
Key idea
A minimum spanning tree adds the cheapest edges that avoid cycles until all vertices connect; on $n$ vertices it always has exactly $n - 1$ edges.
Frequently asked questions
Step-by-step solutions to exam-style questions on Networks in HSC Maths Standard 2, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Standard 2 syllabus for Networks, using the methods and notation expected in the exam.