A bus, a train, and a plane depart from city A to city B at the same time. If I take the bus, which has an average speed of 100 $\mathrm{km} / \mathrm{h}$, I will arrive in city B at 8 PM. If I take the train, which has an average speed of $300 \mathrm{~km} / \mathrm{h}$, I will arrive in city B at 2 PM. What will be the arrival time of the plane if its average speed is $900 \mathrm{~km} / \mathrm{h}$?
Let $d$ be the distance between the two cities and $h$ be the common departure time of the bus, the train, and the airplane. Since distance $=$ speed $\times$ time, we have $d=100 \times(20-h)$ and $d=300 \times(14-h)$. Therefore, $100 \times(20-h)=300 \times(14-h)$, from which $h=11$. Thus, the distance between the two cities is $d=100 \times(20-11)=900 \mathrm{~km}$. Therefore, the airplane takes 1 hour from city $\mathrm{A}$ to city $\mathrm{B}$ and, thus, the airplane arrives at 12 o'clock.