We model the situation by a system of two homogeneous linear differential equations with constant coefficients.
Let $x=x(t)$ be the quantity, in grams, of salt in tank A at time $t$, and $y=y(t)$ the quantity of salt in tank B at time $t$.
We are given the initial conditions $x(0)=60$ and $y(0)=0$.
We now examine the flow of salt in and out of tank A. Brine is flowing out to B at the rate of $20$ liters per minute. At any time $t$, the amount of salt in a liter of water is $x/200$, so salt is flowing out at the rate $(20/200)x$. Also, salty (after a while) water is flowing in from B at rate $5$ liters per minute. Thus salt is flowing into A at the rate $(5/200)y$. The above out-in analysis for the salt can be written as the differential equation
$$\frac{dx}{dt}=-\frac{20}{200}x +\frac{5}{200}y \qquad\qquad\text{(Equation $1$)}$$
Now we do a similar analysis for tank B. Salt is flowing in from A at the rate of $(20/100)x$. Salt is flowing out to A at the rate $(5/200)y$, and to the outlet pipe at the rate $(15/200)y$, for a combined out rate of $(20/200)y$. This yields the differential equation
$$\frac{dy}{dt}=\frac{20}{200}x -\frac{20}{200}y \qquad\qquad\text{(Equation $2$)}$$
We run through the solution process. Let $M$ be the matrix
$$\begin{pmatrix} -\frac{20}{200}& \frac{5}{200}\\ \frac{20}{200}& -\frac{20}{200}
\end{pmatrix}$$
We first find the eigenvalues and associated eigenvectors of the matrix $M$. Standard calculation shows that the eigenvalues are $-1/20$ and $-3/20$. The vector $(1,2)$ (written as a column vector) is an eigenvector for eigenvalue $-1/20$; the vector $(1,-2)$, again written as a column vector, is an eigenvector for eigenvalue $-3/20$.
Theory tells us that the general solution $(x,y)$ is given by
$$(x,y)=Ce^{-t/20}(1,2)+De^{-3t/20}(1,-2)$$
where $C$ and $D$ are constants.
Less compactly,
$$x=Ce^{-t/20}+De^{-3t/20} \qquad \text{and}\qquad y=2Ce^{-t/20}-2De^{-3t/20}.$$
From the fact that $x(0)=60$ we get $C+D=60$. From $y(0)=0$ we get $2C-2D=0$. So $C=D=30$, and therefore
$$x=30e^{-t/20}+30e^{-3t/20} \qquad \text{and}\qquad y=60e^{-t/20}-60e^{-3t/20}.$$
Now we know everything, so we can answer any question. We were asked when there are equal percentages of salt in both tanks. The tanks are of equal size, so equal percentage happens when the amounts of salt are equal, that is, at the time $t$ when $x=y$. Thus
$$30e^{-t/20}+30e^{-3t/20}=60e^{-t/20}-60e^{-3t/20}.$$
This simplifies to $3e^{-3t/20}=e^{-t/20}$, then to $e^{t/10}=3$. Take the natural logarithm of both sides: $t=10\ln 3\approx 10.98$.
It really just is a simple flow in minus flow out, after attention is paid to the units.
$400\,\frac{\rm cm^3}{\rm s}$ = $0.0004\,\frac{\rm m^3}{\rm s}$ and, since the base has area $1\,\frac{\rm m^2}{\rm s}$, the water pumped in at any given moment increases the height by $.04\,\rm cm$.
Now analyze similarly for the outflow and you have the differential equation.
Best Answer
The volume $V$ of a sperical cap of height $h$ in a sphere of radius $r$ is $$V=\pi \left(h^2r-\frac{h^3}{3}\right).$$ The flow satisfies the differential equation $$\frac{dV}{dt}=-S\cdot v(t)$$ where $S$ is the area of the hole at the bottom and $v(t)=0.6 \sqrt{2gh(t)}$ the velocity of the flow through the hole. Hence, we obtain $$\frac{dV}{dt}=\pi \left(h^2r-\frac{h^3}{3}\right)'=\pi(2hr-h^2)h'=-0.6 S\sqrt{2gh}.$$ Now we separate the variables and integrate $$\pi\int_0^{r} \frac{2hr-h^2}{\sqrt{h}}\,dh=0.6 S\sqrt{2g}\cdot T \Rightarrow T=\frac{14\pi r^{5/2}}{15\cdot 0.6 S\sqrt{2g}}$$ where we used the conditions $h(0)=r$ (the hemisphere is full) and $h(T)=0$ (the hemisphere is empty).