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....???

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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?

CPhill Sep 20, 2017

#1**+3 **

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

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

the rate of the "first" boat = r_{1} , the rate of the second boat = r_{2 } ,

and the time it takes for the boats to meet the first time = t_{1}

time = distance / rate

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

t_{1} = (w - 720) / r_{1}

And the second boat goes 720 yards.

t_{1} = 720 / r_{2}

(w - 720) / r_{1} = 720 / r_{2} → r_{1} = (w - 720) / (720 / r_{2})

Then, t_{2} 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.

t_{2} = (720 + w - 400) / r_{1} = (w + 320) / r_{1}

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

t_{2} = (w - 720 + 400) / r_{2} = (w - 320) / r_{2}

(w + 320) / r_{1} = (w - 320) / r_{2} → r_{1} = (w + 320) / ( (w - 320) / r_{2 })

(w - 720) / (720 / r_{2}) = (w + 320) / ( (w - 320) / r_{2 })

(w - 720) * (r_{2} / 720) = (w + 320) * (r_{2} / (w - 320) ) Divide both sides by r_{2} .

(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

#2**+2 **

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.....!!!!!

CPhill Sep 20, 2017

#4**+1 **

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