2. A production team in a factory is tasked with producing a batch of parts. When each worker works at their original position, it takes 9 hours to complete the production task. If the positions of workers A and B are swapped, with the production efficiency of other workers remaining unchanged, the task can be completed one hour earlier; if the positions of workers C and D are swapped, with the production efficiency of other workers remaining unchanged, the task can also be completed one hour earlier. Question: If the positions of A and B, as well as C and D, are swapped simultaneously, with the production efficiency of other workers remaining unchanged, how many minutes earlier can the production task be completed?
2.【Solution】Divide the total task into 72 parts. Originally, $\overline{9}=8$ parts were completed per hour, and each part took $\overline{8}=$ 7.5 minutes.
After A and B are swapped, $\frac{72}{8}=9$ parts are completed per hour, which is 1 part more than before. Since the efficiency of other workers remains unchanged,

this additional part is the extra work done by A and B.
Similarly, after C and D are swapped, they also complete 1 additional part per hour. After both swaps, A and B, as well as C and D, each complete 1 additional part per hour, so the entire group completes $8+1+1=10$ (parts) per hour, meaning each part takes $\frac{60}{10}=6$ (minutes) to complete. Therefore, for each part of the task, the time is reduced by $7.5-6=1.5$ (minutes). For 72 parts, the total time saved is $72 \times 1.5=108$ (minutes).