Question 88201
Imagine that when the tank is full it contains F gallons of water. Since the inlet pipe will
fill the tank in 45 minutes, each minute it will put 1/45 of F into the tank. So in t
minutes, the amount of water that will be added to the tank will be (1/45)*F*t.
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Similarly, since the outlet pipe will empty the tank in 30 minutes, each minute it will let
1/30 of F out of the tank. And in t minutes the amount of water that will flow out of the
tank is (1/30)*F*t
.
Now we can write an equation.  We start with a full tank that contains F gallons. To that 
we add water at the amount of (1/45)*F*t and subtract water at the amount of (1/30)*F*t. 
Eventually the amount of water in the tank will be zero. So the equation becomes:
.
F + (1/45)*F*t - (1/30)*F*t = 0
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Notice now that we can eliminate F by dividing all the terms in this equation (both sides)
by F to get:
.
1 + (1/45)*t - (1/30)*t = 0
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Get rid of the 1 on the left side by subtracting 1 from both sides and the equation
becomes:
.
(1/45)*t - (1/30)*t = -1
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Next you can eliminate the denominators by multiplying all the terms on both sides by
90 to get:
.
2t - 3t = -90
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combine the two terms on the left side any you get:
.
-t = -90
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And multiply both sides by -1 to get:
.
t = 90
.
This result tells you that if you start with a full tank and both and inlet and outlet
valves open such that the inlet would fill the tank in 45 minutes and the outlet would empty
it in 30 minutes, the tank will be empty in 90 minutes.
.
You can sort of see this.  If you start with the full tank, in 90 minutes the inlet valve
(which fills the tank each 45 minutes) would add 2 more tanks of water, for a total of 3 tanks.
.
Meanwhile, the outlet valve which empties a tankful in 30 minutes would empty the 3 full
tanks in 90 minutes. Therefore the tank would be dry in 90 minutes.
.
Hope this helps you to understand the problem.