Question 216558
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So what you are saying is that you don't have enough time to do the work required to research what you want, but we have plenty of time to repeat time and effort previously spent.  I guess you must feel the value of your time is much greater than that of any of us.


If spout B were closed, then spout A would fill the 64 gallon drum in 32 hours.  That means that spout A fills *[tex \Large \frac{1}{32}] of the tank per hour.  We don't know what fraction of the tank spout B empties per hour, so let's just call it *[tex \Large \frac{1}{x}].  We want the tank to fill at a rate of *[tex \Large \frac{1}{96}] per hour.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{32}\ -\ \frac{1}{x}\ =\ \frac{1}{96}]


Just solve for *[tex \Large x] which will be the number of hours it would take spout B to empty a full tank if spout A were closed.  Divide *[tex \Large x] into 64 to get the gallons per hour rate for spout B.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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