Two boats on opposite sides of a river head towards each other at different speeds. When they pass each other the first time they are 700 yards from one shoreline. They continue to the opposite shoreline, turn around, and move towards each other again. When they pass the second time they are 300 yards from the other shoreline.

How wide is the river? (Assume both boats travel at a constant speed and ignore factors such as turn-around time and the current of the river).

Units of time are arbitrary, so set the speed of one boat as 1 and the other as s (which is then just the ratio of their speeds). Let the speed of 1 apply to the boat that has traveled 700 yards less than the river's width at the first meeting, and the speed of s apply to the one that has traveled 700 yards at that time. Call the width x, which will be our final answer.

At the first time:
(x-700)/1 = 700/s
At the second time:
(2x-300)/1 = (x+300)/s

Dividing one equation by the other, and cross-multiplying (or just cross multiplying to begin with):
x²-400x-210000=1400x-210000
or
x²-1800x=0

Ignoring the x=0 solution, we get x=1800 yards.

The ratio of the speeds, by the way, is 7/11.

Posted by Charlie on 2003-11-13 14:53:18