SOLUTION: A hose can fill a swimming pool in 12 hours. Another hose needs 6 more hours to fill the pool than the two hoses combined+How long would it take the second hose to fill the pool?

Question 1094128: A hose can fill a swimming pool in 12 hours. Another hose needs 6 more hours to fill the pool than the two hoses combined+How long would it take the second hose to fill the pool?
Found 2 solutions by greenestamps, Gentle Phill:
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!

Let x be the number of hours the second hose takes to fill the pool by itself. The fraction of the pool that the first hose fills in 1 hour is 1/12; the fraction the second fills in 1 hour is 1/x.

The fraction that the two hoses together fill in 1 hour is

So the number of hours it takes the two hoses together to fill the pool is

The second hose alone needs 6 hours more than that to fill the pool by itself:

The second hose takes 12 hours to fill the pool by itself.

Answer by Gentle Phill(18) (Show Source): You can put this solution on YOUR website!
Let the 1st hose be A and the 2nd one, B
.
Since A can fill 1 pool in 12 hours
.
A(rate) = 1pool/12hrs
.
Simply put:
A = 1/12
.
B(rate) = 1pool/xhrs
.
Simply put:
B = 1/x
.
(A+B) = 1pool/yhrs
.
(A+B) = 1/y
.
(1/12)+(1/x) = 1/y
.
y = 12x/(x+12) ... Eqn 1
.
Just so that you dont get confused;
~ x = time for which only B fills the pool
~ y = time for which both A & B fills the pool.
.
But y = x-6hrs
.
Simply put:
y = x-6
.
Substitute into eqn 1:
x-6 = 12x/(x+12)
.
(x-6)(x+12) = 12x
.
x2+6x-72 = 12x
.
x2+6x-12x-72 = 0
.
x(x+6)-12(x+6) = 0
.
(x-12)(x+6) = 0
.
x = (12, -6)
.
Since the time can not be a negative value, required x = 12
.
Thus, time for which B fills the pool = 12hrs (same as A)
.
Upon evaluation both will fill in same pool within 6hrs which is 6hrs less than only B or A.
.
.
.
Your friend,
Francis.
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