If Alice can make a trip of m miles in t hours, at what rate must she make the return trip, given that it is possible to do so, so that the average speed for the whole journey will be k miles per hour?
We’ve got a lot of variables already, but we go ahead and let x represent the rate for the return trip.
Then the usual chart:
| d = | r | t | |
| go | m | m/t | t |
| come | m | x | m/x |
| avg. | 2m | k | t + m/x |
Hmm: What if 2m = kt? Well, suppose the distance m = 10 miles and the time t = 2 hours, so the “going” rate is 10/2 or 5 miles per hour. Then 2(10) = k · 2 → the desired average is 10 mi/hr. But going 2m = 20 miles at an average speed of 10 mi/hr will take 2 hours, and going already took 2 hours, so there is no time to get back. Thus, we can never ask that the average speed be double the “going” rate.