Posted by AJA on Tuesday, June 21, 2011 at 3:28pm.
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.
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