A tank was filled with water at the rate of 1.5 gal/min. By 10 a.m., it contained 50 gallons several minutes late the water going into the tank was turned off, the plug was pulled, and the tank was drained at the rate 2.3 gal/min. The tank was empty by 11:08 a.m. How many minutes did it take to drain the tank?

Question

anonymous · Answer

this has something to do with systems of equations...

anonymous · Answer

http://answers.yahoo.com/question/index?qid=20100123230636AAbGg74 Let x be the elapsed time in minutes between 10:00 AM and that point in time when the plug was pulled. The volume of water placed in the tank is equal to the water volume removed. Water volume is the product of the pumping/draining rate and the elapsed time in minutes. 50 + 1.5 x = 2.3 (68 - x) where 68 minutes is the difference between 11:08 AM and 10:00 AM. Now solve for x.

anonymous · Answer

so the answer is 28? @robtobey

anonymous · Answer

Yes.