Two boats are crossing the river from opposite sides.
When they first meet, they are 720 feet from the near shore.
When they reach the opposite shores, they stop for 10 minutes
and cross the river again, but this time they meet 400 ft from the
far shore.
How wide is the river?
Let the width of the river be w.
Let the speed of the first boat be f.
Let the speed of the second boat be s.
Time of first meeting is 720/f = (w - 720)/s
f reaches far shore at w/f
waits 10 minutes and heads back
s reaches near shore at w/s
waits 10 minutes and heads back
time of second meeting is
w/f + 10 + 400/f = w/s + 10 + (w-400)/s
- - - - That's the answer - - - - here's the full story.
The 10 minute delay is irrelevant, since both of them take it,
so we'll ignore it.
We can also use any absolute speeds we want, since we
don't have any specific time information. It's just the relative
speeds that matter.
The ratio of speeds of the two boats is 9 : 13.
The faster boat is the one coming from the far shore.
Let's use 104 feet per minute and 72 feet per minute.
Then the first meeting takes place after 10 minutes,
720 feet from the near shore and 1040 feet from the far shore.
The faster boat reaches the near shore after 16.92 minutes.
The slower boat reaches the far shore after 24.44 minutes.
Ignoring the 10 minute delay,
the fast boat turns around and travels 1360 feet in 13.08 minutes,
total time elapsed = 16.92 + 13.08 = 30 minutes,
the slow boat turns around and travels 400 feet in 5.56 minutes,
total time elapsed = 24.44 + 5.56 = 30 minutes!
Increasing or decreasing the absolute speeds won't matter,
except for the actual numbers of minutes.
All the ratios will be preserved.
720/(w-720) = (400 + w)/(2w - 400) 