Swimming pool

Let's take a differential equation approach! Let $V_{0}$ be the volume of the pool. Also, let $\dot{V_{in}(t)} = \frac{V_{0}}{5}$ and $\dot{V_{out}(t)} = \frac{V_{0}}{6}$ be the intake and outtake flow rates for this pool. If we attempt to fill the pool while simultaneously draining it, then we have:

$\dot{V(t)} =\dot{V_{in}(t)} - \dot{V_{out}(t)} = \frac{V_{0}}{5} -\frac{V_{0}}{6} = \frac{V_{0}}{30}$

and we can solve the ODE:

$\dot{V(t)} = \frac{V_{0}}{30}, V(0) = 0 \Rightarrow V(t) = \frac{V_{0}}{30}t$ .

If $T$ is the time to fill the entire pool, then $V_{0} = \frac{V_{0}}{30} \cdot T \Rightarrow \boxed{T= 30 hrs.}$

In one (1) hour, 1/5 of the pool is filled (no drain) In one (1) hour, 1/6 of the pool is drained (no hose) In one (1) hour, 1/5-1/6 = 1/30 of the pool is filled with water. So that it takes 1/30 x 30 hours to fill the pool. 30 hours x 60 ( minutes in a hour ) =1800 min.