A man travels from City A to City B at a speed of 60 km/h, then from City B to City C at 90 km/h, and finally returns directly from City C to City A at 72 km/h. If the distances between A and B and between B and C are equal, with B located between A and C, find his average speed for the entire journey.
72 km/hr
Let the distance AB = BC = \(d\). Since B lies between A and C, the direct distance CA \(= 2d\).
Time for A to B \(= \dfrac{d}{60}\), time for B to C \(= \dfrac{d}{90}\), time for C to A \(= \dfrac{2d}{72}\).
Total time \(= \dfrac{d}{60} + \dfrac{d}{90} + \dfrac{2d}{72} = \dfrac{6d+4d+10d}{360} = \dfrac{20d}{360} = \dfrac{d}{18}\).
Total distance \(= d + d + 2d = 4d\).
Average speed \(= \dfrac{4d}{d/18} = 72\) km/hr.
Hence, his average speed for the entire journey is 72 km/hr.
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