Algebra
posted by Jimmy .
There is a water tank with a drain of an unknown size that is always open. There is a series of input pipes above it which are all the same size. When 4 pipes are on, it takes 6 hours to fill. When 3 pipes are on, it takes 10 hours to fill. How many pipes would it take to fill the tank in 3 hours?

if an input pipe could fill the tank in x hours, and the drain could empty it in y hours with no input,
4/x  1/y = 1/6
3/x  1/y = 1/10
x=15,y=10
so, if n pipes fill the tank in 3 hours,
n/15  1/10 = 1/3
n = 13/2
Hmmm. I was expecting an integer. 
Could you help explain how T can put the number of pipes over the time to solve this?? I'm afraid I don't understand.
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