Here is a problem I've always liked...it was devised by one of the great American "puzzleists" of all time, Sam Lloyd


It has been posted on the forum before, but not in a while.....maybe we have some  new people who would like to try their hand at solving it....???




What is the exact width of the Hudson River?

Two ferry boats start moving at the same instant from 
opposite sides of the Hudson River, one boat going fron 
New York to Jersey City and the other going from Jersey 
City to New York.  One boat is faster than the other, 
so they meet at a point 720 yards from the nearest shore.

After arring at their destinations, each boat remains ten 
minutes in the slip to change passengers; then it starts 
on its return trip.  The boats again meet at a point 400 
yards from the other shore.  What is the exact width of 
the river? 



cool cool cool

CPhill  Sep 20, 2017

5+0 Answers


Ah ha!!! I think I found a way laugh (Though maybe not the best way)


Let's call the width of the river = w ,

the rate of the "first" boat = r1  , the rate of the second boat = r ,

and the time it takes for the boats to meet the first time = t1


time = distance / rate


Let's say the first boat is the one that goes  w - 720 yards. So...


t1  =  (w - 720) / r1


And the second boat goes  720 yards.


t1  =  720 / r2


(w - 720) / r1  =  720 / r2   →   r1  =  (w - 720) / (720 / r2)


Then,  t2  is the time it takes the boats to meet again. So the first boat must travel the remaining 720 yards and then it must go  w - 400  yards to be 400 from the other side.


t2  =  (720 + w - 400) / r1  =  (w + 320) / r1


And the second boat must go the remaining  w - 720 yards and then it must go 400 yards.


t2  =  (w - 720 + 400) / r2  =  (w - 320) / r2


(w + 320) / r1  =  (w - 320) / r2   →   r1  =  (w + 320) / ( (w - 320) / r)


(w - 720) / (720 / r2)  =  (w + 320) / ( (w - 320) / r)

(w - 720) * (r2 / 720)  =  (w + 320) * (r2 / (w - 320) )       Divide both sides by  r2 .

(w - 720) / 720  =  (w + 320) / (w - 320)

Then I used WA to solve this because I am lazy...here


So... I get that the width is  1760  yards....which is exactly 1 mile! 

hectictar  Sep 20, 2017

GOOD JOB , hectictar....that's exactly the correct answer  !!!


BTW  - I solved this several years ago  [ after thinking about it for a LONG time] and used an algebraic approach like you did.....I remember that  Lloyd's solution was rather elegant - no algebra involved - , but I can't remember how he did it.....!!!!!




cool cool cool

CPhill  Sep 20, 2017
edited by CPhill  Sep 20, 2017

Yay!! Thanks!


It does seem like there would be a more elegant way to do it.... smiley

hectictar  Sep 20, 2017

CPhill: Is this solution close to Sam Lloyd's solution?
When the ferry boats meet at point "X" they are 720 yards from one shore.  The combined distance that both have traveled is equal to the width of the river. When they reach the opposite shore, the combined distance is equal to twice the width of the river.  On the return trip they meet at point "Z" after traveling a combined distance of three times the width of the river, so each boat has gone three times as far as they had gone when they first met. At the first meeting, one boat had gone 720 yards, so when it reaches "Z" it must have gone three times that distance, or 2160 yards.  So, this distance is 400 yards more than the river's width, so all the math work we have to do is to subtract 400 from 2,160 to get the river's width.  It is 1,760 yards which is exactly one mile.
The amount of time each ship spent at the landing does not affect the problem.

Guest Sep 21, 2017

Thanks, guest.....that's exactly the solution that Lloyd proposed.....I just couldn't remember how he did it.........it's much more elegant than an algebraic method, but....as they say "any port in a storm".....!!!!!



cool cool cool

CPhill  Sep 23, 2017

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