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Problem of the Week
Problem D and Solution
Layover Between the Trips

Problem

A plane travels from Calgary, AB to Grande Prairie, AB. The total flight time, including takeoff and landing, is \(1\) hour and \(40\) minutes. The return flight takes the same route and time. The average speed for these two flights is \(500\) km/h.

After a brief layover in Grande Prairie, the average speed of this entire round trip (including the two flights and the layover in between) becomes \(425\) km/h. How long was the layover?

Solution

Let \(t\) be the length of the layover, in hours.

The plane travels from Calgary to Grande Prairie in \(1\) hour \(40\) minutes at a speed of \(500\) km/h. Using the formula distance \(=\) speed \(\times\) time, the distance from Calgary to Grande Prairie must be \(500 \frac{\text{km}}{\text{h}} \times 1 \frac{2}{3}\ \text{h} = 500\times \frac{5}{3} = \frac{2500}{3}\) km.

Therefore, for the two-way trip, the plane travels \(2\times \frac{2500}{3} = \frac{5000}{3}\) km.

The length of time of the entire two-way trip is the time of the two flights plus the layover time. Therefore, the total length of time of the trip is \(\frac{5}{3} + \frac{5}{3} + t = \frac{10}{3} + t\) hours.

Since the average speed of the entire two-way trip is \(425\) km/h, using the formula distance \(=\) speed \(\times\) time, we have \[\begin{aligned} \frac{5000}{3} & = 425 \times \left(\frac{10}{3} + t\right) \\ \frac{10}{3} + t &= \frac{5000}{3 \times 425}\\ t &= \frac{200}{51} - \frac{10}{3}\\ &= \frac{200}{51} - \frac{170}{51}\\ &= \frac{10}{17} \end{aligned}\] Therefore, the layover was \(\dfrac{10}{17}\) hours, or approximately \(35\) minutes.