Journey of Two Trains

As nobody has posted any solution, I am giving first hint.

Hint 1: As both trains meet 10th time, the faster train has to complete 9 one-side trip to meet the another train 10th time.

Note: I hope, many viewers will be able to find solution with the above hint. If I don't get solution in another 12 hours, I shall post second hint.

Somehow I missed this one! Here goes nothing:

Extend the tracks so that A has a 15 mile turnaround at each city and that B has a 10 mile turnaround at each city. Now force each train to run constantly, without stops, in Gomez Addams fashion.

Call a full trip the time it takes a train to go from newly facing one city to newly facing the other city (after the turnaround). A's full trip = 241 miles and B's full trip = 236 miles.

To meet the slower train B ten times, the faster train A will complete 9 full trips, then make his tenth encounter with B somewhere in the midst of his tenth full trip. This is not a trivial deduction, but we can unpack it in another post.

When A finishes his ninth full trip, he's traveled 9*241 miles, which will take him 9*241/60 hours, or 36.15 hours.

As A begins his tenth full trip, we know that B has traveled for the same time as A, or 36.15 hours. B has traveled at 40 mph, so B's total distance = 36.15 * 40 = 1446. Each of B's trips is 236 miles, and 1446/236 has remainder 30, so B is 30 miles into his seventh trip.

Since B is 30 miles into the 226 miles that separate the cities and A has just begun his tenth full trip, the trains have (226 - 30) miles between them. Their joint rate is 100 mph, so they'll meet in (226 - 30)/100 hours or 1.96 hours. 36.15 + 1.96 = 38.11, so that looks like our answer.