Question 1196842
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let x = # of hours to fill the tank; then 1/x is the fraction of the tank that is filled in 1 hour
let y = # of hours to drain the tank; then 1/y is the fraction of the tank that is drained in 1 hour<br>
The tank is filled and then drained in 9 hours, so<br>
(1) {{{x+y = 9}}}<br>
When the tank is being drained at the same time it is being filled, the fraction of the tank that gets filled in 1 hour is 1/x - 1/y.  In that situation, the tank gets filled in 20 hours, so 1/20 of the tank is getting filled each hour:<br>
(2) {{{1/x-1/y=1/20}}}<br>
There are two equations in x and y that you can solve by many different methods.  Perhaps this....<br>
{{{y=9-x}}}
{{{1/x-1/(9-x)=1/20}}}<br>
Multiply through by the least common multiple of the denominators, which is {{{20(x)(9-x)}}}:<br>
{{{20(9-x)-20x=x(9-x)}}}
{{{180-20x-20x=9x-x^2}}}
{{{x^2-49x+180=0}}}
{{{(x-45)(x-4)=0}}}<br>
The time to fill the tank is either 45 hours or 4 hours -- but the 45 hours is not consistent with the given information.  So<br>
ANSWER: The number of hours to fill the tank with the drain closed is x = 4<br>
Note that is a lot of ugly algebra to solve a problem that can be solved mentally by playing with numbers a bit:<br>
1/20 = 1/4 - 1/5; and 4+5 = 9....<br>