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April 18, 2015

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

- math -
**Reiny**, Tuesday, June 21, 2011 at 4:30pmLet 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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