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Saw this question online.

  1. Aug 17, 2012 #1
    The amount of water in a tank doubles every minute. If the tank was full at the 1 hour mark, when was the tank half full?
    It's not homework, I'm just trying to get the actual (reasoned) answer..
    Here is the answer I argued (though, I'm not sure):
    2^x = [(2^59)/2]
    2^x = 2^58
    x = 58
    But, almost everyone argues:
    If it doubles every minute, isn't it half full the minute before it's full? ie. 59

    Pretty sure I'm incorrect, I just want to know for sure..
    Last edited: Aug 17, 2012
  2. jcsd
  3. Aug 17, 2012 #2
    The argument of everyone (not yours) is correct.
    Your working is correct as well, its just that your first equation is wrong.
    Last edited: Aug 17, 2012
  4. Aug 17, 2012 #3


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    Homework Helper

    V = k*2x, where x is in minutes and k is some unknown constant (the volume of the tank at the start of the timer x=0).

    At x=60 the tank is full, therefore, if we denote the full volume of the tank as Vf then Vf=k*260. Now, We want to know when the tank is half full, which is Vf/2

    Clearly, Vf/2 = k*260/2 = k*259

    If we compare this to the original equation, x=59 minutes as you would expect.

    I know this explanation is long-winded, but I hope it got the point across as to how you should have set up the equation.
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