Problem description:

At 6 o'clock (), the minute hand is behind the hour hand 100 degrees.

At 5 o'clock (), the minute hand is 50 degrees beyond the hour hand.

At 9 o'clock (), the angle between the minute hand and the hour hand is 150 degrees.

At 7 o'clock (), the hour hand and the minute hand are symmetrical about the line of 6 words and 12 words.

Analysis:

According to: the hour hand moves 30 degrees per hour, each branch is 0.5 degrees, the minute hand moves 360 degrees per hour, and each branch is 6 degrees.

At 1 and 6 o'clock, if the hour hand is on 6, the minute hand is on 12, and 12 is 0 degrees, the hour hand is ahead of the minute hand 180 degrees.

(180-100)/(6-0.5) =160/1=14 and 6/1/kloc-0.

That is 6: 00, 14 and 6/ 1 1.

2.5 o'clock, if the hour hand is on the 5th, the point where the minute hand is at 12 and 12 is 0 degrees, and the hour hand is ahead of the minute hand 150 degrees. Then, according to the question:

(150+50)/(6-0.5) = 400/1= 36 and 4/ 1 1.

That is, 5: 36 and 4/ 1 1.

3.9 o'clock, hour hand at 9, minute hand at 12 and 12 at 0 degrees, the hour hand is 270 degrees ahead of the minute hand. There are two situations:

① (150-90)/(6-0.5) =120/1=1and 9/1/kloc-

That is 9: 00, 1 1 and 9/ 1 1.

② (270-150)/(6-0.5) = 240/1= 21and 9/1.

That is 9: 00, 2 1 and 9/ 1 1.

At 4.7 o'clock, when the hour hand is at 7, the minute hand is at 12, and the point of 12 is 0 degrees, the hour hand leads the minute hand by 2 10 degrees, and the equation is obtained:

2 10+0.5x- 180 = 180-6x

6.5x=330

X=660/ 13=50 and 10/ 13.

That is 7: 50 and 10/ 13.