How long will it take to fill a water tank with two inlet pipes and one outlet? E11/40. 
I can see here that I work the fill rate out as:
$\large\frac13 + \frac14 - \frac18 =$
Overall fill rate of $\large\frac{11}{24}$ tank per hour.
If I multiply $60$ minutes by $\large\frac{24}{11}$ I get the correct result of $131$ minutes, but I can't explain why I calculate it that way...
 A: In one hour you fill $\frac{11}{24}$ of the tank.  So in $\frac{1}{11}$ of an hour you fill one of the $\frac{11}{24}$s of the tank; that is, you fill $\frac{1}{24}$ of the tank in $\frac{1}{11}$ of an hour.  So it will take you $\frac{24}{11}$ of an hour to fill $\frac{24}{24}$ of the tank.  Convert the $\frac{24}{11}$ of an hour to minutes to get your answer. 
A: Let S be the volume of the tank.
t is time
S is a number between 0 and 1 (empty and full)
$S(t) \in [0,1]$


*

*The first inlet pipe


liquid flows at speed $V_1$ (this is constant)
$S(0) = 0$ (empty)
$S(3) = 1$ (full)
$S(t) = V_1 * t + C$ (this comes from assuming constant water speed)
$0 = S(0) = V_1 * 0 + C$
so C = 0
$1 = S(3) = V_1 * 3 + 0$
$V_1 = 1/3$


*

*The second inlet pipe


This is all similar
$V_2 = 1/4$


*

*The Outlet pipe


Again this is all the same except that this is emptying so use a negative sign, the flow is in the opposite direction.
$V_3 = -1/8$
So the overall rate of flow is the sum of these
$V = V_1 + V_2 + V_3$
Let $S_A$ be the volume of the tank when ALL pipes are open
This is $S_A = V * t$
$S_A = 1$ is a full tank
The time it takes for this is 
$S_A / V = t$
or $1/ V = t = 24/11$
