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?

Answer 1

Let time taken by smaller pipe be t hrs
then time takes by larger pipe is t-6 hrs

rate of smaller pipe = 1/t
rate of larger pipe = 1/(t-6)
combined rate = 1/t + 1/(t-6)
= (2t-6)/(t(t-6))

time at combined rate = 1/ [ (2t-6)/(t(t-6)) ]
= t(t-6)/(2t-6)

then t(t-6)/(2t-6) = 4
t^2 - 6t = 8t - 48
t^2 - 14t + 48 = 0
(t-8)(t-6) = 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.