A woman from a village started at 10:31 and arrived at village $B$ at 13:43, traveling at a constant speed. On the same day, a man started from $B$ at 9:13 on the same route and arrived at $A$ at 11:53, also at a constant speed. They met at a bridge along the way, and after passing each other, the woman left the bridge 1 minute later than the man. When did they reach the bridge and where are the endpoints of the bridge?
I. solution: The woman completed the entire journey in 192 minutes, and the man in 160 minutes. Since both walk at a constant speed, the ratio of the time required by the woman and the man to cover the same distance is always

$$
\frac{192}{160}=\frac{6}{5}
$$

Let $t$ be the time, in minutes, it takes the man to cross the bridge. According to the problem, the woman takes $t+1$ minutes to cross the bridge, and thus

$$
\frac{t+1}{t}=1+\frac{1}{t}=\frac{6}{5}, \quad t=5 \text { minutes }
$$

Choosing the length of the bridge as the unit of distance, the total length of the journey (calculated from the man's travel time) is

$$
160: 5=32 \text { bridge lengths. }
$$

The woman and the man together covered 31 bridge lengths before reaching the bridge. If the woman's distance from this was $x$ bridge lengths, then, since the woman covers one bridge length in 6 minutes, she reached the bridge in $6x$ time. The man started 78 minutes earlier than the woman and walked $(31-x) 5$ minutes to reach the bridge. Since they arrived at the bridge at the same time, we have

$$
\begin{gathered}
6 x=(31-x) 5-78 \\
x=7
\end{gathered}
$$

Accordingly, the woman (and simultaneously the man) reached the bridge 42 minutes after the woman started, or at 11:13 AM, from the time the woman started.

The bridge's $A$-side pier is 7 bridge lengths from $A$, and thus the other pier is 24 bridge lengths from $B$.

II. solution: The man completes the entire journey in 160 minutes and has been walking for 78 minutes when the woman starts, so he has already covered

$$
\frac{78}{160}=\frac{39}{80}
$$

of the journey, and together they still have to cover $\frac{41}{80}$ of the journey until they meet.

Since both walk at a constant speed, the ratio of their speeds is

$$
\frac{192}{160}=\frac{6}{5}
$$

in other words, the distance the man covers in 5 minutes, the woman covers in 6 minutes. From this, it immediately follows that they cross the bridge in 5 and 6 minutes, respectively. The above ratio can also be interpreted as the woman covering $\frac{5}{11}$ and the man $\frac{6}{11}$ of the distance they cover together in the same time. Thus, they meet at

$$
\frac{5}{11} \cdot \frac{41}{80}=\frac{41}{176}
$$

of the total distance from $A$, and this will be at $\frac{5}{11}$ of the bridge from $A$. The length of the bridge (calculated from the man's travel time) is

$$
\frac{5}{160}=\frac{1}{32}
$$

of the total journey, so the end of the bridge closer to $A$ is

$$
\frac{41}{176}-\frac{5}{11} \cdot \frac{1}{32}=\frac{77}{352}=\frac{7}{32}
$$

of the total distance from $A$, and the far end is at the quarter distance from $A$.

The man starting from $B$ covers the distance to the bridge, which is $\frac{3}{4}$ of the total journey, in $\frac{3}{4} \cdot 160=120$ minutes, or 2 hours, so both arrive at the bridge at 11:13 AM.