math
posted by AJA on .
Filling a tank. Two pipes are connected to the same tank. Working together they can fill the tank in 4 hrs. The larger pipe working alone can fill the tank 6 hrs less than the smaller one, HOw long would the smaller one take, working alone, to fill the tank?

Let time taken by smaller pipe be t hrs
then time takes by larger pipe is t6 hrs
rate of smaller pipe = 1/t
rate of larger pipe = 1/(t6)
combined rate = 1/t + 1/(t6)
= (2t6)/(t(t6))
time at combined rate = 1/ [ (2t6)/(t(t6)) ]
= t(t6)/(2t6)
then t(t6)/(2t6) = 4
t^2  6t = 8t  48
t^2  14t + 48 = 0
(t8)(t6) = 0
t = 8 or t = 6 , but t = 6 would make the rate of the larger pipe undefined, so we reject that answer
So it would take 8 hours to fill with only the smaller pipe.