| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Discrete (Further Paper 3 Discrete) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Route Inspection |
| Type | Route inspection with time constraint |
| Difficulty | Standard +0.8 This is a standard Chinese Postman Problem requiring identification of odd vertices, pairing them optimally, and calculating the route length with a unit conversion. The 'one side only' detail is a minor conceptual twist but doesn't significantly complicate the solution. Typical Further Maths Decision question requiring systematic application of an algorithm rather than novel insight. |
| Spec | 7.04e Route inspection: Chinese postman, pairing odd nodes |
| Answer | Marks | Guidance |
|---|---|---|
| Odd nodes: \(B, C, F, G\); Shortest distances: \(B\)–\(C\): 160, \(B\)–\(F\): 60, \(B\)–\(G\): 235, \(C\)–\(F\): 220, \(C\)–\(G\): 75, \(F\)–\(G\): 295 | M1 | Sets up as route inspection problem, identifies \(B, C, F, G\) as odd-degree nodes (PI) |
| Pairings: \((B\)–\(C)(F\)–\(G) = 455\); \((B\)–\(F)(C\)–\(G) = 135^*\); \((B\)–\(G)(C\)–\(F) = 455\) | A1 | All three correct totals for pairs of shortest distances |
| Distance of all streets in town centre \(= 1090\) m | B1 | |
| Minimum distance \(= 1090 + 135 = 1225\) m; shortest possible time \(= \frac{1225}{4800} \times 60 = 15\) minutes | M1 | |
| 15 minutes (CAO, condone lack of units) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| The traffic warden may need to stop to issue fines, reducing average speed, causing an increase in time to complete the walk | E1 | Plausible reason given with consequences explained |
## Question 6(a):
| Odd nodes: $B, C, F, G$; Shortest distances: $B$–$C$: 160, $B$–$F$: 60, $B$–$G$: 235, $C$–$F$: 220, $C$–$G$: 75, $F$–$G$: 295 | M1 | Sets up as route inspection problem, identifies $B, C, F, G$ as odd-degree nodes (PI) |
| Pairings: $(B$–$C)(F$–$G) = 455$; $(B$–$F)(C$–$G) = 135^*$; $(B$–$G)(C$–$F) = 455$ | A1 | All three correct totals for pairs of shortest distances |
| Distance of all streets in town centre $= 1090$ m | B1 | |
| Minimum distance $= 1090 + 135 = 1225$ m; shortest possible time $= \frac{1225}{4800} \times 60 = 15$ minutes | M1 | |
| 15 minutes (CAO, condone lack of units) | A1 | |
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## Question 6(b):
| The traffic warden may need to stop to issue fines, reducing average speed, causing an increase in time to complete the walk | E1 | Plausible reason given with consequences explained |6 A council wants to monitor how long cars are being parked for in short-stay parking bays in a town centre. They employ a traffic warden to walk along the streets in the town centre and issue fines to drivers who park for longer than the stated time.
The network below shows streets in the town centre which have short-stay parking bays. Each node represents a street corner and the weight of each arc represents the length, in metres, of the street.
The short-stay parking bays are positioned along only one side of each street.\\
\includegraphics[max width=\textwidth, alt={}, center]{22f11ce2-8d07-4f51-9326-b578d1e454f9-08_483_1108_733_466}
The council assumes that the traffic warden will walk at an average speed of 4.8 kilometres per hour when not issuing fines.
6
\begin{enumerate}[label=(\alph*)]
\item To monitor all of the parking bays, the traffic warden needs to walk along every street in the town centre at least once, starting and finishing at the same street corner.
Find the shortest possible time, to the nearest minute, it can take the traffic warden to monitor all of the parking bays.
Fully justify your answer.\\
6
\item Explain why the actual time for the traffic warden to walk along every street in the town centre at least once may be different to the value found in part (a).
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Discrete 2019 Q6 [6]}}