Pipes A and C can fill an empty cistern in 32 and 48 hours, respectively while Pipe B can drain the filled cistern in 24 hours. If the three pipes are turned on together when the cistern is empty, how many hours will it take for the cistern to be $$\frac{2}{3}$$ full?
In 1 hour pipe A can fill 1/32, pipe C 1/48 and pipe B can drain 1/24 of the cistern.
When A, BÂ and C are opened : 1/32 + 1/48-1/24 = 1/96
So cistern full in 96 hour
for $$\frac{2}{3}of pipe =\frac{2}{3}\times96$$
= $$64$$ hours
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