Can you solve this problem in your head?
This is a neat little problem that I came across. When looked at the right way, it requires just some simple arithmetic and no algebra.
Two ferries traveling across a river at constant speeds meet 700 feet from one of the river banks. They continue to the opposite end, turn around and meet again 400 feet from the opposite river bank. How wide is the river?
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6 Answers
At the time of the first meeting, if the ferries left from opposite shores, the combined distance travelled will be exactly one width of the river.
At the time of the second meeting, each boat will have travelled one complete width of the river, plus together they will have travelled a third width.
The time taken before the ferries meet again must be three times the time taken to meet the first time as they will have travelled exactly three times the distance at the same speed.
This means the ferry that had travelled 700 feet at the time of the first meeting has had time to travel 2,100 feet by the time of the second meeting.
This ferry is now 400 feet from the shore on its way back to where it started so the river’s width must be 2,100 – 400 = 1,700 feet.
The way I’ve set it up in my head, both ferries have traveled for t time. In that time, one ferry has gone 700’ + 700’ + ( x – 400)’. The other ferry has traveled x’ + ( x – 400)’.
I think the problem is that even assuming “constant speeds”, we don’t know if both ferries are traveling at the same speed. For that reason I think this is an insoluble problem with the information given. If we knew that the ferries were traveling at the same speed, then it would be a simple matter to equate both of the above “distance traveled” equations to figure the value of “x”, where “x” = the remaining width of the river over 700’.
@flutherother , Well done. I was hoping that it would take just a little bit longer for someone to get the answer.
I’d seen something similar before. It is tempting to use algebra but that would complicate rather than simplify.
The obvious variation on this problem is if the 700 foot and 400 foot distances are from the same river bank. @flutherother ’ s method can still be used, but now it requires some simple algebra.
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