SOLUTION: One pipe can fill a storage tank in five hours less time than a smaller second pipe can, and together they can fill it in six hours. How long would it take the larger pipe to fill

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Question 370569: One pipe can fill a storage tank in five hours less time than a smaller second pipe can, and together they can fill it in six hours. How long would it take the larger pipe to fill it?
Found 2 solutions by ewatrrr, mananth:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi,

Let x represent the time it takes for the smaller pipe to fill the tank

then (x-5) would be the time for the larger pipe

Question states:  Per hr being the equalizer

1/x + 1/(x-5) = 1/6hr

Multiplying each term on both sides of the equation by 6x(x-5) so as all denominators = 1

6(x-5) + 6x = x(x-5)

6x - 30 +6x = x^2 -5x

x^2 -17x +30 = 0

factoring

(x -15)(x-2) = 0

(x-2) = 0    x = 2 (cannot use)

(x -15)= 0   x = 15hr , the time it takes for the smaller pipe to fill the tank

the time would it take the larger pipe to fill it, is 10hr.  (15-5)

checking our answer

1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6

Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
let second pipe take x hours to fill
it will do 1/x of the job in 1 hour.
..
First pipe takes 5 hour less than second pipe.
x-5 hours
..
it can do 1/(x-5) of the job in 1hour.
... Together they do the job in 6 hours..
together they do 1/6 of the job in hour.
...
1/x + 1/(x-5 ) = 1/6
..
LCD = x(x-5)
(x-5+x)/(x(x-5))=1/6
(2x-5)/x(x-5) = 1/6
6(2x-5)=x(x-5)
12x-30=x^2-5x
x^2-5x-12x+30=0
x^2-17x+30=0
x^2-15x-2x+30=0
x(x-15)-2(x-15)=0
(x-15)(x-2)=0
x= 2 OR 15 hours
...
m.ananth@hotmail.ca