On the way out, a car covers the first half of a 240 km route at 60 km/h and the second half at 40 km/h. On the way back it runs at 60 km/h for half the time taken and at 40 km/h for the other half of the time. The average speed on the return trip exceeds that on the outward trip by how many km/h?

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Accepted Answer

2. Equal distances: the average is the harmonic mean 2 × 60 × 40/(60 + 40) = 48 km/h, because the slower half eats up more of the time. Equal times: the average is the plain mean (60 + 40)/2 = 50 km/h. The return trip is faster by 50 − 48 = 2 km/h. (Using (60 + 40)/2 for both trips gives 0 and misses the whole point of the contrast.)

On the way out, a car covers the first half of a 240 km route at 60 km/h and the second half at 40 km/h. On the way back it runs at 60 km/h for half the time taken and at 40 km/h for the other half of the time. The average speed on the return trip exceeds that on the outward trip by how many km/h?

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