SOLUTION: One pipe can fill a tank in 6 hours less than another. Together they can fill the tank in 4 hours. How long would it take each alone to fill the tank?

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Question 262073: One pipe can fill a tank in 6 hours less than another. Together they can fill the tank in 4 hours. How long would it take each alone to fill the tank?
Found 2 solutions by palanisamy, richwmiller:
Answer by palanisamy(496) About Me  (Show Source):
You can put this solution on YOUR website!
Let the pipe A fill the tank in x hours.
Then the pipe B fill the tank in (x-6) hours.
In one hour, the pipe A fills 1/x part of the tank.
And,in one hour the pipe B fills 1/(x-6) part of the tank.
So both pipes together fill 1/x+1/(x-6) part of the tank in one hour.
In one hour, both pipes together fill [x+x-6]/x(x-6) part of the tank.
Time taken to fill the tank is x(x-6)/(2x-6)= 4
x^2-6x = 8x-24
x^2-14x+24 = 0
(x-2)(x-12) = 0
x = 2 or 12
X cannot be equal to 2.
So, x = 12
So the pipe A fill the tank in 12 hours.
Then the pipe B fill the tank in (12-6)= 6 hours.

Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
4/(x-6)+4/x=1
4x+4(x-6)=(x-6)*x
8x-24=x^2-6x
0=x^2-14x+24
(x-2)(x-12)=0
x=2 or x=12
we can't use x=2 since we can't have -4 hours
4/12+4/6=1
2/6+4/6=1
true
x=12 hours
one in 12 hours and one in 6 hours