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?
:
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/t}}} + {{{4/((t-6))}}} = 1
:
Multiply by t(t-6)
t(t-6)*{{{4/t}}} + t(t-6)*{{{4/((t-6))}}} = 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)
{{{4/12}}} + {{{4/6)}}} = 1
{{{1/3}}} + {{{2/3)}}} = 1
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