5 replies to this topic

[ThunderCloud]

This one isn't too hard, but is a fun exercise.

 

Two motor boats began speeding toward one another at the same time from opposite shores of a river. Upon reaching the opposing shore, each motor boat immediately reversed course and headed back to his original shore. In so doing, the motor boats passed each other twice. The first time they crossed, they were 700 feet from one shore of the river; the second time, they were 300 feet from the opposite shore. Assuming each boat traveled with constant speed, and neglecting any influence of river current, how wide is the river?

Edited by ThunderCloud, 25 February 2013 - 01:39 AM.

[BMAD]

Spoiler for best guess

1800 ft.

[Debasis]

Spoiler for

they wre 1400 ft apart...

Edited by Debasis, 26 February 2013 - 12:18 PM.

[Debasis]

Spoiler for best guess

1800 ft.

 

why?

[ThunderCloud]

 

Spoiler for best guess

1800 ft.

 

why?

 

Spoiler for Detailed Solution

Call the width of the river w. Then from the moment the boats began moving to the moment the boats crossed for the first time, boat "A" traveled 700 feet and boat "B" traveled (w - 700) feet. Since they completed their respective distances in the same amount of time, and each boat is traveling with a constant speed (say Va for boat "A" and Vb for boat "B"), we have that 700 / Va = (w - 700) / Vb. By similar reasoning, from the moment the boats first crossed to the moment they crossed a second time, boat "A" completed a distance of (w - 700) + 300, while boat "B" completed a distance of 700 + (w - 300), and so we have that (w - 400) / Va = (w + 400) / Vb.

 

Dividing one equation by the other cancels out the terms Va and Vb, leaving: 700 / (w - 400) = (w - 700) / (w + 400). With a few algebraic manipulations, this simplifies to w*(w - 1800) = 0, from which the answer is apparent.

[Debasis]

 

Spoiler for best guess

1800 ft.

 

why?

 

Spoiler for Detailed Solution

Call the width of the river w. Then from the moment the boats began moving to the moment the boats crossed for the first time, boat "A" traveled 700 feet and boat "B" traveled (w - 700) feet. Since they completed their respective distances in the same amount of time, and each boat is traveling with a constant speed (say Va for boat "A" and Vb for boat "B"), we have that 700 / Va = (w - 700) / Vb. By similar reasoning, from the moment the boats first crossed to the moment they crossed a second time, boat "A" completed a distance of (w - 700) + 300, while boat "B" completed a distance of 700 + (w - 300), and so we have that (w - 400) / Va = (w + 400) / Vb.

 

Dividing one equation by the other cancels out the terms Va and Vb, leaving: 700 / (w - 400) = (w - 700) / (w + 400). With a few algebraic manipulations, this simplifies to w*(w - 1800) = 0, from which the answer is apparent.

Thanks!