How far apart are the ferryboats when they start their return trips?

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The discussion revolves around a mathematical problem involving two ferryboats traveling across a river. They start from opposite shores and meet at a point 720 yards from one shore. After a 20-minute wait, they return and meet again 400 yards from the opposite shore. The key to solving the problem lies in establishing the relationship between the distances covered by the boats at the two meeting points. The calculation reveals that the width of the river is 1760 yards, derived from the equation that relates the distances traveled by each boat. The simplicity of the solution is acknowledged, with participants expressing relief at arriving at the correct answer.
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post your answers in white :biggrin: . no pen or paper allowed.

Two ferryboats start at the same instant from opposite sides of a river, traveling across the water on routes at right angles to the shores. Each travels at a constant speed, but one is faster than the other. They pass at a point 720 yards from the nearest shore. both boats remain in their slips for 20 minutes before starting back. On the return trips they meet 400 yards from the other shore.

How wide is the river?I didn't get the right answer and then I felt like an idiot because the answer is really easy :frown: .
 
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The answer is:
1760[/color]
yards
 
moe darklight said:
How wide is the river?

Well,
at the two moments the ferryboats meet, the relation between the distances covered by them is the same:
(w - a) / a = (2w - b) / (w + b)
So,
w = 3a - b = 3*720 - 400
w = 1760 yards

:smile:
 
oops, totally forgot about this thread— yep, both right :biggrin: .
 
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