Question 5
A pilot flies a mail plane from its base, W, to four stations, J, K, L and M. The plane lands once at each station and then returns to W. The table shows the flight distance (km) between locations. Two distances are not shown.
| Flight | W–J | W–K | W–L | W–M | J–K | J–L | J–M | K–L | K–M | L–M |
| Distance (km) | 465 | 120 | 345 | 194 | 435 | 347 | 272 | 272 |
The table shows the coordinates of W, J and K.
| Location | Latitude | Longitude |
|---|---|---|
| W | 25.0° S | 141.0° E |
| J | 25.0° S | 143.5° E |
| K | 21°30′ S | 143.5° E |
The plane flies at an average speed of 260 km/h and spends 40 minutes on the ground at each station. The pilot leaves W at 9:15 am and flies the shortest possible round trip.
Determine the earliest time the pilot can arrive back at W.
Measure two flights, try every cycle, then add the time
W–J runs along a parallel and J–K along a meridian. All 12 round trips are totalled, the shortest is chosen, and its flying and ground times give the arrival back at W.
Pick a round trip and time it
Pick any of the 12 round trips from W. The map draws it as a thick dashed loop, and the readouts give its total, flying time and arrival time to the nearest minute.
Thin solid lines: every flight, in km. Thick dashed loop: the round trip you picked. L and M are placed for the picture only.
Orange blocks: flying. White blocks: 40 min on the ground.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
W and J are on the same parallel of latitude (25° S), so use the distance along a parallel.
$$D = 111.2\cos\theta \times \text{angular distance}$$
$$\text{angular distance} = 143.5^\circ - 141.0^\circ = 2.5^\circ$$
$$D_{\text{WJ}} = 111.2\cos 25^\circ \times 2.5 = 251.95\text{ km}$$
W and J are both on the 25° S parallel, so use the parallel form with $\theta = 25^\circ$ and the difference in longitude.
Check the calculator is in degree mode (D on the display), then key 111.2 × cos 25 × 2.5.
An equivalent value, for example 252 km, also earns this mark.
Marker: correctly determines the distance between W and J
J and K are on the same meridian (143.5° E). Convert the minutes to decimal degrees first.
$$21^\circ 30' = 21.5^\circ \qquad \text{angular distance} = 25.0^\circ - 21.5^\circ = 3.5^\circ$$
$$D = 111.2 \times \text{angular distance}$$
$$D_{\text{JK}} = 111.2 \times 3.5 = 389.2\text{ km}$$
J and K are both on the 143.5° E meridian, so every degree of latitude is 111.2 km. Convert the minutes first $(60' = 1^\circ)$.
An equivalent value, for example 389 km, also earns this mark.
Marker: correctly determines the distance between J and K
List every round trip from W that visits each station once (a Hamiltonian cycle), grouped by the two stations next to W, and total each one.
| Stations next to W | Hamiltonian cycle | Total distance (km) |
|---|---|---|
| J, K | W–J–L–M–K–W | 251.95 + 194 + 272 + 272 + 465 = 1454.95 |
| J, K | W–J–M–L–K–W | 251.95 + 435 + 272 + 347 + 465 = 1770.95 |
| J, L | W–J–K–M–L–W | 251.95 + 389.2 + 272 + 272 + 120 = 1305.15 |
| J, L | W–J–M–K–L–W | 251.95 + 435 + 272 + 347 + 120 = 1425.95 |
| J, M | W–J–K–L–M–W | 251.95 + 389.2 + 347 + 272 + 345 = 1605.15 |
| J, M | W–J–L–K–M–W | 251.95 + 194 + 347 + 272 + 345 = 1409.95 |
| K, L | W–K–J–M–L–W | 465 + 389.2 + 435 + 272 + 120 = 1681.2 |
| K, L | W–K–M–J–L–W | 465 + 272 + 435 + 194 + 120 = 1486 |
| K, M | W–K–J–L–M–W | 465 + 389.2 + 194 + 272 + 345 = 1665.2 |
| K, M | W–K–L–J–M–W | 465 + 347 + 194 + 435 + 345 = 1786 |
| L, M | W–L–J–K–M–W | 120 + 194 + 389.2 + 272 + 345 = 1320.2 |
| L, M | W–L–K–J–M–W | 120 + 347 + 389.2 + 435 + 345 = 1636.2 |
The plane visits each station once and returns to W, so the trip is a Hamiltonian cycle. With four stations there are $\dfrac{4!}{2} = 12$ cycles (each can be flown in reverse for the same total), so list them all, grouped by the two stations next to W. A shortcut such as nearest neighbour does not guarantee the shortest cycle, so it cannot identify the shortest round trip.
Alternative reasoning, for example a weighted graph with the totals of the possible Hamiltonian cycles, also earns this mark.
Marker: provides mathematical reasoning or working to identify the shortest round trip
The shortest round trip is the cycle with the smallest total.
Shortest round trip: W–J–K–M–L–W (or W–L–M–K–J–W), 1305.15 km
Follow-through applies from an error in earlier working.
Marker: determines the total distance of the shortest round trip
Divide the distance by the speed, then convert the decimal part of an hour to minutes.
$$\text{time} = \frac{\text{distance}}{\text{speed}} = \frac{1305.15}{260} = 5.0198\text{ h}$$
$$0.0198 \times 60 = 1.19\text{ min} \qquad \text{flying time} = 5\text{ h }1\text{ min}$$
The speed applies to flying only, so divide the distance by the speed, then convert the decimal part of an hour to minutes. Times are stated to the nearest minute, and 1.19 min is 1 min 11 s, so round down to 1 min (the exact arrival is 4:56:11 pm).
With 5.0198 on the display, the degrees-minutes-seconds key (°′″) shows 5°1′11″, which reads as 5 h 1 min 11 s.
This mark may be implied by subsequent working.
Marker: determines the total flying time
Add the time on the ground at the four stations, then add the total time to 9:15 am.
$$\text{ground time} = 4 \times 40 = 160\text{ min} = 2\text{ h }40\text{ min}$$
$$\text{total time} = 5\text{ h }1\text{ min} + 2\text{ h }40\text{ min} = 7\text{ h }41\text{ min}$$
$$9{:}15\text{ am} + 7\text{ h }41\text{ min} = 4{:}56\text{ pm}$$
The plane lands at the four stations only. The trip ends on arrival at W, so there is no ground time at W.
An equivalent time, for example 4:57 pm or 1656 hours, also earns this mark.
A response that includes time on the ground at W does not earn this mark.
Marker: determines the earliest arrival time back at W
Set the response out in order and end with a statement in context.
The pilot flies the shortest round trip, W–J–K–M–L–W, a distance of 1305.15 km. The flying time is 5 h 1 min and the time on the ground is 2 h 40 min, so the earliest time the pilot can arrive back at W is 4:56 pm.
The response must run in order (missing distances, all cycles, shortest route, time) with units, and end with a statement in context.
Logical organisation includes clear flow of the solution using appropriate mathematical terminology, symbols, conventions and representations.
Marker: shows logical organisation
If I had to guess where you might lose marks…
- Leaving out $\cos 25^\circ$ for W–J gives $111.2 \times 2.5 = 278$ km, and radian mode gives 275.55 km. Either loses Mark 1. W–J–K–M–L–W then totals 1331.2 km (or 1328.75 km), so W–L–J–K–M–W (1320.2 km) looks shortest and the arrival time becomes about 5:00 pm.
- Reading $21^\circ 30'$ as $21.3^\circ$: the angular distance becomes $3.7^\circ$ and J–K becomes 411.44 km. This loses Mark 2. The shortest total becomes 1327.39 km and the arrival time 5:01 pm.
- Using nearest neighbour from W instead of listing all 12 cycles: it gives W–L–J–K–M–W, 1320.2 km, which is not the shortest. This loses Mark 4 and the arrival time becomes about 5:00 pm.
- Adding 40 min on the ground at W as well ($5 \times 40 = 200$ min) gives 5:36 pm. This loses Mark 6.
- Reading 5.0198 h as 5 h 19.8 min instead of 5 h 1 min gives 5:15 pm. This loses Mark 5.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it