Question 426390: Two pipes are connected to the same tank. Working together, they can fill the tank in 4 hours. The larger pipe working alone can fill the tank in 6 hrs less than the smaller one. How long would the smaller one take, working alone to fill the tank?
I tried making the large tank X and the small tank x+6
then came up with
x=6x=4 but that does not work I am stuck on where to start
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! "The larger pipe working alone can fill the tank in 6 hrs less than the smaller one" ___ L = s - 6
each pipe fills a FRACTION of the tank , based on the individual rates
___ the fractions add to one , the WHOLE tank
(4 / L) + (4 / s) = 1
substituting for the value of L ___ [4 / (s - 6)] + (4 / s) = 1
multiplying by [s(s-6)] ___ 4s + 4s - 24 = s^2 - 6s
0 = s^2 - 14s + 24
factoring ___ 0 = (s - 2)(s - 12)
s - 2 = 0 ___ s = 2 ___ this value is not realistic because L is 6 less than s , and L is not negative
s - 12 = 0 ___ s = 12
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