Answer
False
Work Step by Step
The distance, $d$, from the garage to Terminal B is $3950\text{ ft}$. The time, $t$, it takes to travel from the garage to Terminal B is
$$
t=60\text{ s}+45\text{ s}=105\text{ s}
.$$
Using $d=rt$, then
$$\begin{aligned}
3950&=r(105)
\\
r&\approx37.62
.\end{aligned}$$
Let $r_1$ be the speed from the garage to Terminal A. Then the speed from Terminal A to Terminal B is $r_1+8.1$. Then,
$$\begin{aligned}
r_1+(r_1+8.1)&=r
\\
r_1+(r_1+8.1)&\approx 37.62
\\
2r_1&\approx 29.52
\\
r_1&\approx 14.76
.\end{aligned}$$
Therefore, the speed from the garage to Terminal A is approximately $14.76\text{ ft/s}$.
Converting $14.76\text{ ft/s}$ to $\text{mi/h}$ results in
$$
\frac{14.76\text{ ft}}{\text{s}}\cdot\frac{1\text{ mi}}{5280\text{ ft}}\cdot\frac{3600\text{ s}}{1\text{ h}}\approx\frac{10.06\text{ mi}}{\text{h}}
.$$
Therefore, the average speed from the garage to Terminal A is approximately $10.06\text{ mi/h}$. Hence, the claim that the monorail's average speed is greater than $25\text{ mi/h}$ is FALSE.
