| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Route Inspection |
| Type | Optimal starting/finishing vertices |
| Difficulty | Challenging +1.2 This is a Chinese Postman Problem application requiring students to identify Eulerian paths/circuits, calculate odd-degree vertices, and find minimum matching. While it requires understanding of graph theory concepts from Further Maths discrete, the problem is relatively standard with clear structure: part (a) is semi-Eulerian (start/end at B), part (b) finds optimal Eulerian circuit. The context is straightforward and the techniques are direct applications of taught algorithms rather than requiring novel insight. |
| Spec | 7.02g Eulerian graphs: vertex degrees and traversability7.04e Route inspection: Chinese postman, pairing odd nodes |
| Answer | Marks |
|---|---|
| 7(a) | Sets up a model by identifying |
| Answer | Marks | Guidance |
|---|---|---|
| identifying the four odd nodes | 3.3 | M1 |
Total length = 240 mm + 44 mm
| Answer | Marks | Guidance |
|---|---|---|
| between odd nodes | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| three pairs | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing/incorrect units | 1.1b | A1 |
| Total | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(b) | Refines the model to semi- |
| Answer | Marks | Guidance |
|---|---|---|
| PI | 3.5c | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing/incorrect units | 3.2a | A1 |
| Total | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 7(c) | States a plausible reason for | |
| starting from B | 3.5b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Total | 1 | |
| Question total | 7 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Sets up a model by identifying
the problem as a route
inspection problem and
identifying the four odd nodes | 3.3 | M1 | Odd nodes are A, C, G and I
Shortest lengths between odd
nodes
A–C: 30 G–I: 26
A–G: 22 C–I: 22
A–I: 43 C–G: 43
Pairings
(A–C)(G–I) = 30 + 26 = 56
(A–G)(C–I) = 22 + 22 = 44*
(A–I)(C–G) = 43 + 43 = 86
44 mm is the shortest length to be
repeated
Total length = 240 mm + 44 mm
= 284 mm
Uses the model to find at least
four correct shortest distances
between odd nodes | 3.4 | M1
Finds the correct minimum pair
of shortest distances from the
three pairs | 1.1b | A1
Determines correctly the
minimum length
CSO
Condone missing/incorrect units | 1.1b | A1
Total | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | Refines the model to semi-
Eulerian by identifying suitable
start and end nodes
PI | 3.5c | M1 | Start at A and end at G
240mm + 22 mm = 262 mm
Finds the correct minimum
length
Condone missing/incorrect units | 3.2a | A1
Total | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c) ---
7(c) | States a plausible reason for
starting from B | 3.5b | B1 | The pendant would be symmetrical
if starting from B
Total | 1
Question total | 7
Q | Marking instructions | AO | Marks | Typical solutionA jeweller is making pendants.
Each pendant is made by bending a single, continuous strand of wire.
Each pendant has the same design as shown below.
\includegraphics{figure_7}
The lengths on the diagram are in millimetres. The sum of these lengths is 240 mm
As the jeweller does not cut the wire, some sections require a double length of wire.
\begin{enumerate}[label=(\alph*)]
\item The jeweller makes a pendant by starting and finishing at $B$
Find the minimum length of the strand of wire that the jeweller needs to make the pendant.
Fully justify your answer.
[4 marks]
\item The jeweller makes another pendant of the same design.
Find the minimum possible length for the strand of wire that the jeweller would need.
[2 marks]
\item By considering the differences between the pendants in part (a) and part (b), state one reason why the jeweller may prefer the pendant in part (a).
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2021 Q7 [7]}}