SOLUTION: Two pipes are connected to a tank. When working together, the two pipes can fill the tank in 4 hours. The larger pipe, working alone, can fill the tank in 6 less hours than the sm

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Question 195111: Two pipes are connected to a tank. When working together, the two pipes can fill the tank in 4 hours. The larger pipe, working alone, can fill the tank in 6 less hours than the smaller one. How long would the smaller one take to fill the tank alone?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Two pipes are connected to a tank. When working together, the two pipes can fill
the tank in 4 hours. The larger pipe, working alone, can fill the tank in 6 less
hours than the smaller one. How long would the smaller one take to fill the tank alone?
:
Let t = time for the small pipe to fill the tank alone
Then
(t-6) = time for the large pipe to fill the tank alone
:
Let the full tank = 1
:
Each tank will fill a fraction of the tank. The two fractions = 1
:
4%2Ft + 4%2F%28%28t-6%29%29 = 1
:
Multiply by t(t-6)
t(t-6)*4%2Ft + t(t-6)*4%2F%28%28t-6%29%29 = t(t-6)(1)
Cancel out the denominators and you have:
4(t-6) + 4t = t(t-6)
:
4t - 24 + 4t = t^2 - 6t
Arrange as a quadratic equation
0 = t^2 - 6t - 8t + 24
:
t^2 - 14t + 24 = 0
Factor this to
(t-2)(t-12) = 0
The only reasonable solution
t = 12 hrs, the small pipe alone
:
:
Check solution in original equation, (large pipe alone:12-6 = 6hr)
4%2F12 + 4%2F6%29 = 1
1%2F3 + 2%2F3%29 = 1
:
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