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Question 98662: A drain can empty a pool in 6 hours. Another drain takes 3 more hours to empty the pool than the two drains combined. How long would it take the second drain to empty the pool?
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
First drain empties the pool at the rate of (1/6) pool per hour
Let x=amount of time it takes second drain to empty the pool
Then second drain empties the pool at the rate of (1/x) pools per hour
Combined the two drains empty the pool at the rate of (1/6+1/x or (x+6)/6x)pools per hour
If t=time it take both drains to empty the pool, then:
((x+6)/6x)t=1
(note: the 1 denotes 1 pool)
Multiply both sides by 6x
(x+6)t=6x divide both sides by (x+6)
t=6x/(x+6) hours--------------time it takes both drains to empty the pool
Now we are told that the second drain takes 3 hours longer than this so our equation to solve is:
x=(6x/(x+6))+3 multiply each term by x+6
x(x+6)=6x+3(x+6) get rid of parens
x^2+6x=6x+3x+18 subtract 6x, 3x and 18 from both sides
x^2+6x-6x-3x-18=6x-6x+3x-3x+18-18 combine like terms
x^2-3x-18=0-------quadratic in standard form and it can be factored:
(x-6)(x+3)=0
x-6=0
x=6 hours--------time it takes second drain to empty the pool
and
x+3=0
x=-3 hours discount the negative value, since time is positive in this case
CK
Time it takes both drains to empty pool:
6x/(x+6)=36/12=3 hours
It takes second drain three hours longer than both combined
6-3=3
3=3
Hope this helps----ptaylor
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