In order to let everyone better master the problem-solving methods of primary school application problems, I sorted out the clock application problems in the second grade of primary school. Please see:

The significance of the clock problem in the second grade of primary school is to study the relationship between the hour hand and the minute hand on the clock surface, such as the coincidence of the two hands, the verticality of the two hands, the alignment of the two hands and the angle between the two hands of 60 degrees. The clock problem can be compared with the catch-up problem.

The speed of the minute hand is 12 times that of the hour hand.

The speed difference between them is1112.

Usually it is treated as a catch-up problem, and it can also be calculated as a differential multiple problem.

The idea and method of solving problems can be modified to "chasing problems" and the formula can be used directly.

Example 1 How many minutes after the hour hand points to 4 o'clock, when the hour hand coincides with the minute hand?

The clock face is divided into 60 squares a week, and the minute hand walks one square every minute and 60 squares every hour; The hour hand moves 5 squares per hour, and 5/60 =112 squares per minute. The minute hand travels more distance per minute than the hour hand (1-112) =11/2 square. At 4 o'clock sharp, the hour hand is in front, the minute hand is behind, and the two hands are 20 squares apart. therefore

The time for the minute hand to catch up with the hour hand is 20 ≈ (1-112) ≈ 22 (minutes).

A: After 22 minutes, the hour hand coincides with the minute hand.

Between four o'clock and five o'clock, when do the hour hand and the minute hand form a right angle?

There are 60 squares on the clock face, and its 1/4 is 15 squares, so the difference between two right-angled hands is 15 squares (including the case where the minute hand is 15 squares before or after the hour hand). At four o'clock sharp, the minute hand is behind the hour hand (5×4). If the minute hand is at right angles to it, it will travel more than the hour hand (5× 4- 15). If the minute hand is at right angles to the hour hand, it will travel more than the hour hand (5× 4+ 15). According to 1 min, the minute hand moves more than the hour hand (1-112), we can find out the time when the two hands are at right angles.

(5× 4-15) ÷ (1-12) ≈ 6 (point)

(5× 4+15) ÷ (1-112) ≈ 38 (point)

A: At 4: 06 and 4: 38, the hands are at right angles.

When do the hour and minute hands coincide between six and seven?

At six o'clock, the minute hand is behind the hour hand (5×6). If the minute hand coincides with the hour hand, it must catch up with the hour hand. This is actually a catch-up problem.

(5× 6) ÷ (1-112) ≈ 33 (point)

A: At 6: 33, the minute hand coincides with the hour hand.

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