I like Sam Lloyd's solution to this problem.
Since the faster of the two ferry boats will cross the halfway point first, it follows that the 'nearest bank' is the one from which the slower boat has left and in which case the slower boat will have travelled 720 yards when they meet.
Also, the sum of the distances travelled by both boats (when they meet), will equal the width of the river.
When they next meet, their combined total distances travelled will be three times the width of the river, and in which case the slower of the boats will have travelled 3*720 = 2160 yards.
Since this is the width of the river plus 400 yards, it follows that the width of the river is
2160 - 400 = 1760 yards.
- Bertie
