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one pipe can fill a tank in 5 hours. another can empty the tank in 8 hours. If both pipes are left open, how long will it take to fill the tank??

please help me answer this anyone.. haha

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Philip P. | Effective and Affordable Math TutorEffective and Affordable Math Tutor
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Let V =the volume of the tank
The first pipe fills 1/5 of the tank per hour:  Rate 1 = +V/5 
The second pipe empties 1/8 of the tank per hour:  Rate 2 = -V/8
Rate 1 is positive because it's filling the tank.  Rate 2 is negative because it's emptying the tank. We leave each pipe open for a time t, and we want the tank to be full (volume = V) at the end of that time.
(Rate 1)*time + (Rate 2)*time = full tank (V)
(V/5)t + (-V/8)t = V
(1/5 - 1/8)Vt = V   (Factor out V and t)
(3/40)t = 1     (Subtract the fractions, divide both sides by V)
t = 40/3 = 13 1/3 hours (13 hours 20 minutes)
(V/5)(40/3) + (-V/8)(40/3) = (8/3)V - (5/3)V = (3/3)V = V     It fills the tank!


Another equivalent way to do it is to compute the net rate of the two pipes:
Net Rate = Rate 1 + Rate 2 = V(1/5 - 1/8) = V*(3/40)
The pipes fill 3/40 of the tank per hour.  Therefore it will take 40/3 hours to fill the tank.