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Thread: A Boat Trick

  1. #1
    Silver Lounger
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    A Boat Trick

    Two ferryboats traveling at a constant speed start moving at the same instant from opposite sides of the Imaginary River. One going from Bigsitty to Litllesitty and the other going from Littlesitty to Bigsitty. They pass one another at a point 720 yards from the Bigsitty shore.

    After arriving at their respective destinations, each boat spends precisely 00:17:30 (17 minutes and 30 seconds) at the opposite shore to change passengers before switching directions. On the return trip, the two boats meet at a point 400 yards from the Littlesitty shore.

    What is the width of the river?
    - Ricky

  2. #2
    Plutonium Lounger
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    Re: A Boat Trick

    <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>1760 yards = 1 mile. The "precisely 00:17:30" is a red herring, the only thing that is important is that the time is the same.

    Denote the width of the river (in yards) by w, the speed (in yards per unit of time) of the first ferry by s1, and that of the second by s2. (In reality, the speed won't be constant, but I think the puzzle can't be solved if you have to take different accellerations into account.)

    When the boats pass each other the first time, the first one has crossed a distance of 720 yards, the second one of w-720 yards. The time taken by both is the same, so 720/s1 = (w-720)/s2. This can easily be transformed to 720(s1+s2) = ws1 (equation A)

    When the boats pass each other the second time, the first one has crossed a total distance of w+400 yards, the second one of w+(w-400) yards. The time taken by both (excluding the time they spent at the shore, which is equal) is the same, so (w+400)/s1 = (2w-400)/s2. This can be transformed to 400(s1+s2) = w(2s1-s2) (equation [img]/forums/images/smilies/cool.gif[/img]

    We can multiply equation A by 5 and equation B by 9 to get 3600(s1+s2) on the left hand side of both. So the right hand sides are equal too: 5ws1 = 9w(2s1-s2). We can drop w from this equation: 5s1 = 18s1-9s2, from which we get 9s2 = 13s1 (equation C, giving the relative speeds)

    Substitute this back into equation A to get 720s1+1040s1 = ws1. This time, we can drop s1, and we have w = 1760.</font color=yellow></span hi>

  3. #3
    Silver Lounger
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    Re: A Boat Trick

    <span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>It was good to see that the "17 minute thing" didn't throw you overboard. I had originally included passenger counts...</font color=yellow></span hi> <img src=/S/grin.gif border=0 alt=grin width=15 height=15>
    - Ricky

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