Tank Half Full Time: Solving the Doubling Water Tank Problem

[Bradyns]

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?
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It's not homework, I'm just trying to get the actual (reasoned) answer..
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Here is the answer I argued (though, I'm not sure):
2^x = [(2^59)/2]
2^x = 2^58
x = 58
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But, almost everyone argues:
If it doubles every minute, isn't it half full the minute before it's full? ie. 59
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Pretty sure I'm incorrect, I just want to know for sure..

 

[Millennial]

The argument of everyone (not yours) is correct.
Your working is correct as well, its just that your first equation is wrong.

 

[Mentallic]

V = k*2^x, 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 V_f then V_f=k*2⁶⁰. Now, We want to know when the tank is half full, which is V_f/2

Clearly, V_f/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.