This problem can also be solved by the relationship of distance: at 6 o'clock, the hour hand is in front and the minute hand is behind. The speed of the hour hand is 112, and the minute hand is1. At this time, the two hands are 30 squares apart. When forming a reverse straight line for the second time, the minute hand is in front and the hour hand is behind, that is, the minute hand exceeds the hour hand by 30 squares. Therefore, the formula: 2× 30 ÷ (1-12) = 60 ÷112 = 720//kloc-0. I wonder if you are satisfied?