Let $L=$Loaded Miles and $U=$Unloaded Miles.
The percentage of loaded to unloaded would be $$100 \times \frac{L}{U+L}$$
For example if (as in your example) $L=150, U=50$ then percentage of loaded miles is
\begin{eqnarray}
100 \times \frac{150}{150+50} &=& 100 \times \frac{150}{200} \\
&=& 100 \times 0.75 \\
&=& 75\%
\end{eqnarray}
This works on the premise you know at least two values from Total Miles, $U$ and $L$.
To solve your question in a general manner, first note there's the required condition
$$x + y = 100 \tag{1}\label{eq1A}$$
Since your equations are symmetric in $x$ and $y$, then WLOG have $x \ge y \implies \lvert x - y \rvert = x - y$. Next, using $120\% = \frac{120}{100} = 1.2$ gives that
$$\begin{equation}\begin{aligned}
\frac{x - y}{(x + y)/2} & = 1.2 \\
\frac{x - y}{100/2} & = 1.2 \\
x - y & = 1.2(50) \\
x - y & = 60
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Now, adding \eqref{eq1A} and \eqref{eq2A} results in
$$2x = 160 \implies x = 80 \tag{3}\label{eq3A}$$
Using this in \eqref{eq1A} then gives
$$80 + y = 100 \implies y = 20 \tag{4}\label{eq4A}$$
Since you already have that $4$ and $16$ give the required difference percentage, here is a bit shorter and easier way to solve the problem. Note that for any $c \neq 0$, we get
$$\begin{equation}\begin{aligned}
\frac{\lvert cx - cy \rvert}{(cx + cy)/2} & = \frac{c\lvert x - y \rvert}{c(x + y)/2} \\
& = \frac{\lvert x - y \rvert}{(x + y)/2}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
i.e., multiplying both $x$ and $y$ by $c$ (and, thus, the sum of them) results in the same value difference percentage. Thus, since $4 + 16 = 20$ and $\frac{100}{20} = 5$, we can use $c = 5$ to directly get that the values of $x$ and $y$ are $4 \times 5 = 20$ and $16 \times 5 = 80$.
Best Answer
Here are two suggestions to show you why your question is not well defined per se. I have suggested two possible curves to model data points between the known extremes at $(65,100)$ and $(93,0)$. Which curve do you prefer and why?