6. (12 points) Person A and Person B walk from location A to location B. B starts at 6:00 AM and walks at a constant speed; A starts at 8:00 AM and also walks at a constant speed. A's speed is 2.5 times that of B, but A needs to rest for half an hour after every half an hour of walking. After A starts, it will take $\qquad$ minutes for A to catch up with B.
【Solution】Solution 1: Assume that A walks 5 meters per hour, and B walks 2 meters per hour. The table is as follows:
\begin{tabular}{|c|c|c|c|c|c|}
\hline Time & A (meters) & B (meters) & Time & A (meters) & B (meters) \\
\hline 0 hours & 0 & 4 & 3 hours & 7.5 & 10 \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|c|c|c|}
\hline 0.5 hours & 2.5 & 5 & 3.5 hours & 10 & 11 \\
\hline 1 hour & 2.5 & 6 & 4 hours & 10 & 12 \\
\hline 1.5 hours & 5 & 7 & 4.5 hours & 12.5 & 13 \\
\hline 2 hours & 5 & 8 & 5 hours & 12.5 & 14 \\
\hline 2.5 hours & 7.5 & 9 & 5.5 hours & 15 & 15 \\
\hline
\end{tabular}

Observing the table, we see that A catches up with B at 5.5 hours (if this time exceeds B's time, the specific catch-up formula should be used to calculate the catch-up time). Solution 2: We can also assume A's speed is $10a$ per hour (A rests, so the actual speed is $5a$ per hour), and B's speed is $4a$ per hour. Therefore, A needs to catch up $8a$. In half an hour, the maximum catch-up is $3a$, so we can subtract $3a$ from the $8a$ to be caught, as A cannot catch up before this (the distance difference is more than $3a$). After that, we start listing every half hour. If it is less than half an hour, we use the catch-up formula. The $5a$ caught before is equivalent to catching $a$ per hour, which can be calculated as $5a \div (5a - 4a) = 5$ (hours). After that, A walks $2a$ in half an hour, and B walks $5a$, plus the remaining $3a$, which is exactly caught up. Therefore, it takes 5.5 hours to catch up, which is 330 minutes.
Thus, the answer is: 330.