A bus normally runs its whole route at a steady 60 km/h. One day, after covering exactly half the route at 60 km/h, it had to crawl through the second half at 40 km/h and so reached the terminus 25 minutes late. The length of the route (in km) is:

For a route of D km the usual time is D/60 h; that day it was D/120 + D/80 = D/48 h. The delay is D/48 − D/60 = D/240 h = 25/60 h, so D = 100 km (usual 100 min, actual 50 + 75 = 125 min). Averaging the two speeds to 50 km/h gives the wrong answer 125 km, since the bus does not spend equal times at the two speeds; applying 40 km/h to the whole route gives 50 km.