2024 Paper 2, Q5
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° S 136° E) | Q | R (20° S 147° E) | |
|---|---|---|---|
| A (20° S 136° 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.
First, find the two missing distances
Two cells in the table are letters. The coordinates tell you how to fill them: A and P share a meridian, and A and R share a parallel of latitude.
Run the Hungarian algorithm
With every distance known, this is a standard 3 × 3 allocation. Reduce rows, reduce columns, count the lines, then allocate.
Earth geometry feeds the matrix
A and P are both at 136° E, so they are on the same great circle. Each degree of latitude between them is 111.2 km.
A and R are both at 20° S. Along a small circle, one degree of longitude is \(111.2\cos\theta\) km, where \(\theta\) is the latitude.
\(x\) and \(y\) each earn a mark. If either is wrong, follow-through still applies to the Hungarian steps.
Move site P
Slide site P north or south along the 136° E meridian. The distance from A changes, and so do all six totals. Once P is within about 5° of A, a different allocation becomes optimal.
A and R are both at 20° S, 11° of longitude apart. Which calculation gives the distance between them?
That is the rule for a great circle. Lines of latitude other than the equator are small circles, so the distance per degree is shorter.
Yes. Along the 20° S parallel, one degree of longitude is \(111.2\cos20^\circ\approx104.5\) km, so \(y\approx1149\) km.
The cosine uses the latitude of the parallel, 20°, and it multiplies the difference in longitude, 11°.
QCAA marking guide · 6 marks
\[\text{angular distance}=28^\circ-20^\circ=8^\circ\]\[D=111.2\times8\approx890\ \text{km}\]
Marker: correctly calculates distance \(x\) in kilometres. Appropriate rounding is accepted.
\[\text{angular distance}=147^\circ-136^\circ=11^\circ\]\[D=111.2\times\cos20^\circ\times11\approx1149\ \text{km}\]
Marker: correctly calculates distance \(y\) in kilometres.
\[\begin{bmatrix}890&600&1149\\445&485&340\\980&1170&770\end{bmatrix}\xrightarrow{\;-600,\ -340,\ -770\;}\begin{bmatrix}290&0&549\\105&145&0\\210&400&0\end{bmatrix}\]
Marker: reduces each row. Columns first is also accepted. Follow-through marks are allowed from here on.
\[\xrightarrow{\;C_1-105\;}\begin{bmatrix}185&0&549\\0&145&0\\105&400&0\end{bmatrix}\]
Marker: reduces each column. Three lines are now needed to cover all zeros, which equals the number of tasks.
Row A has only one zero, under Q. Row C has only one zero, under R. That leaves P for B.
\[\text{A}\to\text{Q},\quad\text{B}\to\text{P},\quad\text{C}\to\text{R}\]
Marker: identifies the optimal allocation for each plane. Listing all six allocations and their totals is also accepted.
\[600+445+770=1815\ \text{km}\]
Send A to Q, B to P and C to R, for a minimum total of 1815 km.
Marker: determines the minimum total distance flown.
Checking against all six
A to Q, B to R, C to P totals 1920 km. That gap of 105 km is the same 105 that was subtracted from column P.
A’s closest site is Q at 600 km, and it goes there. But B does not go to its closest site, R. Sending C to R saves more, because C’s other options are much longer.
Two topics from different units meet here. The Hungarian algorithm is familiar, but the matrix is incomplete, and the missing entries can only be found with Earth geometry. You have to notice that one pair of places shares a longitude and the other shares a latitude, and use a different rule for each. When a matrix has letters in it, look at what else the question gives you before you start reducing.
Question wording and marking-guide steps are from the 2024 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.
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