One, (20 points) Class A and Class B set out from School $A$ simultaneously to a military camp $B$ for military training, which is $75 \mathrm{~km}$ away from the school. The walking speed of Class A students is $4 \mathrm{~km} / \mathrm{h}$, and the walking speed of Class B students is $5 \mathrm{~km} / \mathrm{h}$. The school has a car, which travels at a speed of $40 \mathrm{~km} / \mathrm{h}$ when empty and $20 \mathrm{~km} / \mathrm{h}$ when loaded, and this car can carry only one class of students at a time. Now, it is required that both classes of students arrive at the military camp simultaneously. How much time do they need at least to arrive?
As shown in Figure 7, let the students of Class A travel by car from point $A$ to point $E$, then walk, covering a distance of $a \text{ km}$ by car, and the car returns empty to point $C$; the students of Class B board the car at point $C$, by which time they have already walked $b \text{ km}$.

The time taken by the car from $A \rightarrow E \rightarrow C$ is equal to the time taken by the students of Class B from $A \rightarrow C$; the time taken by the car from $E \rightarrow C \rightarrow B$ is equal to the time taken by the students of Class A from $E \rightarrow B$. Therefore, we have
$$
\left\{\begin{array}{l}
\frac{a}{20}+\frac{a-b}{40}=\frac{b}{5}, \\
\frac{a-b}{40}+\frac{75-b}{20}=\frac{75-a}{4} .
\end{array}\right.
$$

Solving these equations, we get $a=60, b=20$.
Thus, the minimum time required for both classes to reach the military camp simultaneously is
$$
\frac{60}{20}+\frac{15}{4}=6 \frac{3}{4}(h) \text {. }
$$