Clock Problem Formula, Clock Problem Mathematics

30 and 5 11 points.

Answer: After the two hands coincide at 12 o'clock, it will take another 5 and 5 11 minutes to form an angle of 30 degrees for the first time.

At an angle of 30 degrees, the ratio of the angle of the hour hand to the minute hand is 1 to 12, and the angle of the minute hand is x, then the angle of the hour hand is x 12x-x 12=30

The solution is x=360 11

Because every minute that passes, the minute hand turns 6 degrees.

So x is divided by 6

Equal to 60-11 minutes. Thanks a lot.

The hour hand travels 6 degrees per minute, and the minute hand travels 6 degrees per minute.

After the two hands coincide at 12 o'clock, the minute hand is relatively fast, which can be understood as a travel problem, that is, how long does it take for the minute hand to travel 30 degrees more than the hour hand.

After x minutes, the angle between the minute and hour hands is 30 degrees.

30, 60 11 minutes.

Solution: At 7 o'clock, the angle between the two needles is 150 degrees, and the two needles between 7 o'clock and 8 o'clock are in a straight line, then the hour hand is 30 degrees longer than the ratio hand, and the time is set to 7 o'clock after x minutes.

x = 60 for 11 minutes.

When the two hands coincide, if 7 o'clock has passed Y minutes, the minute hand will travel 360-150 = 210 degrees more than the hour hand.

y=420 for 11 min.

So the time taken to solve the problem is: 420 11-60 11=360 11 minutes.

So the starting time is 7:60 11 minutes, and the time is 360 11 minutes.

When the minute hand is in front of the hour hand, the clamp angle is formulated as n*6°-(m*30°+n*, when the minute hand is behind the hour hand, the clamp angle is (m*30°+n*). where n is the minute and m is the hour. The degree is the number obtained by measuring the degree of grandeur, and refers to the standard used for measurement.

A common form of clock problem is clock face chasing. The problem of clock face tracking is usually a problem that studies the position between the hour and minute hands, such as "the coincidence of the minute and hour hands, perpendicular, straight lines, and how many degrees of angle are formed". The hour and minute hands move in the same direction, but the speed and pie are different, similar to the catch-up problem in the travel problem.

The key to solving this problem is to determine the speed or difference of the hour hand obscuring the old lead, the minute hand, or the speed difference. In the specific process of solving the problem, we can use the grid method, when the circumference of the clock face is evenly divided into 60 cells, each grid is called 1 grid. The minute hand travels once an hour, i.e. 60 minutes, while the hour hand travels only 5 minutes per hour, so the minute hand travels 1 minute per minute, and the hour hand travels 1 12 minutes per minute.

The speed difference is 11 to 12 divisions. It is also possible to use the degree method, that is, from an angle point of view, the circumference of the clock face is 360 °, and the minute hand rotates 360 60 degrees per minute, that is, the speed of the minute hand is 6 ° min, and the hour hand turns 360 12 = 30 degrees per hour, so the speed per minute is 30 ° 60, that is. The difference in speed between the minute hand and the hour hand is.

One is 1 minute faster, the other is 3 minutes slower, and the difference is 4 minutes, so it's 1: 3 on the side of the book, 10 o'clock on the fast clock, 9 o'clock on the slow clock, so the standard time is between them.

Yes with fractions Yes two can be used with a fast bell or with a slow bell fast clock: 10-60x1 4 slow clock: 9+60x3 Zen posture 4 The answer is d 9:45

The entire clock face is 360 degrees, with 12 large squares on it, each of which is 30 degrees; The nucleus is 60 cells, each of which is a 6-degree coarse bush.

2. Minute hand speed: 1 small grid per minute, 6 degrees per minute.

3. Hour hand speed: one-twelfth of a small grid per minute, walking degrees per minute.

If you can't use the equation, you can divide the distance by the velocity difference, which is 9V divided by (12V-V) to calculate the time.

Analysis] According to life experience, it is not difficult for us to find that when looking in the mirror, although the image in the mirror is upright, but the left and right sides of the socks are reversed, and there are the following ways to summarize such topics: Method 1: "Reverse reading method" Looking at the picture from the back of the paper, using the conventional reading method, it is easy to read the time at the moment is 3:40. Method 2:

If you read the numbers in the counterclockwise direction, the numbers on the graph are also arranged in the counterclockwise direction from small to large, and it is not difficult to read the time at this time at 3:40. Method 3: "Twelve Deduction Method" After reading the time on the clock according to the regular reading method, subtract this moment from the twelve is the actual time, and the regular reading time is 8:

20, the actual time is 12:00-8:20=3:

40。Method 4: "Symmetry Method" One of the characteristics of plane mirror imaging is that the image and the object are reversed from side to side, so it can be concluded according to the nature of the axisymmetric image, that is, the moment is 3:40 The answer is c.

Comments] This problem belongs to the practical application of the imaging characteristics of the plane mirror, and several common methods for solving the clock problem are listed in the analysis.

There are two cases when the hour and minute hands are in a straight line:

The first case: coincidence.

Analysis: At 8 o'clock sharp, the minute hand lags behind the hour hand 8 5 40 (divisions), the minute hand travels 1 block in 1 minute, and the hour hand moves 5 60 1 12 (divisions in 1 minute), and the problem is transformed into a chase problem. So the minute hand can catch the hour hand 1 (1 12) 11 12 (divisions) in 1 minute, then it takes 40 (11 12) 480 11 (minutes) 43 and 7 11 minutes to catch the hour hand 40 divisions.

Solution: (8 5) [1-(1 12)] 480 11 (min) 43 and 7 11 min.

Answer: At 8:43 and 7 11, the hour and minute hands are in a straight line.

The second case: on the same straight line, but not coincidentally.

Analysis: This situation is divided into two small cases:

1) The minute hand is 180° ahead of the hourly hand

On the clock, each subdivision is 6°. 180 ° is 30 divisions, the minute hand is 40 divisions behind, but also 30 squares ahead, if the hour hand does not move, it takes 70 minutes, not to mention that the hour hand is still moving forward, so this situation is not between 8 o'clock and 9 o'clock.

2) The minute hand is 180° behind the hour hand

180 ° is 30 divisions, then the minute hand is 40 divisions behind at 8 o'clock, this situation requires it to be 30 divisions behind, so the minute hand needs to chase the hour hand 40 30 10 (division), and the pursuit of 10 divisions takes a total of 10 (11 12) 120 11 minutes 11 and 10 11 minutes.

Solution: There are 2 cases when the minute hand lags behind the hour hand

1) The minute hand is 180° ahead of the hourly hand

180 6) 40] [1-(1 12)] = 70 (11 12) 840 11>60 (undesirable, discarded).

2) The minute hand is 180° behind the hour hand

40 (180 6)] 1-(1 12)] = 10 (11 12) 120 11 11 and 10 11 points.

Answer: At 8:10 and 10 11, the minute and hour hands are in a straight line.

The clock issue is, actually"Travel issues"There is no fixed formula for the catch-up problem.

There are a total of 12 large grids around the clock, the hour hand travels 1 large grid 30 degrees per hour, and the instant hand moves 30 60 = degrees per minute;

If the minute hand goes 1 revolution per hour, then the minute hand travels 360 60 = 6 degrees per minute.

Example (1): Question: At what time is 4 o'clock, when the hour hand and minute hand are at right angles for the first time?

Solution: At 4 o'clock, the minute hand falls into 4 large divisions, totaling 120 degrees; If the first time the hour hand is at right angles to the minute hand, the minute hand will have to travel (120-90) degrees more than the hour hand, and the time required is: (120-90) (minutes) = 5 and 5 11 minutes.

A: At 4:60 11 (4:5 and 5 11), the hour and minute hands are at right angles for the first time.

Example (2): At a few points at 2 o'clock, numbers"3"Exactly halfway between the hour and minute hands?

Solution: Set 2 points x minutes when the number"3"Directly between the hour and minute hands, then: the hour hand has gone (degrees, and the minute hand has gone (6x) degrees.

The hour hand is off the numerals"3"The degrees are;

The minute hand is off the numerals"3"The power is: 6x-90

Then the solution is x=240 13=18 and 6 13

Answer: 2:240 13 (i.e. 2:18 and 6:13) at the time of the number"3"Exactly halfway between the hour and minute hands?

The hour hand is 1 12 r h, the minute hand is 1 r h, and the hour and minute hands are half a circle apart at the start.

1/2÷(1-1/12)=6/11h

6 to 11 hours.

Clock Problems The problem of the relationship between the hour and minute hands on the clock face is studied. The clock face is divided into 60 compartments. When the minute hand goes 60 grids, the hour hand goes exactly 5 squares, so the speed of the hour hand is 5 60 = 1 12 of the minute hand, and every 60 (1 5 60) = 65 + 5 11 (minutes) of the minute hand coincides with the celery shed, the clock problem is varied, and there is also a lot of knowledge.

Here's a basic formula: the number of squares to catch up at the initial moment (1 1 12) = the catch-up time (minutes), where 1 1 12 is the number of squares per minute minute hand that is more than the hour hand.

Common clock problems: find the angle between the hour hand and the minute hand at a certain time, the two needles coincide, the two needles are vertical, and the two needles are in a straight line.

It is set between 4 o'clock and 5 o'clock, and it is 4 o'clock x minutes when the hour hand is at an angle of 45 degrees to the minute hand. Rule.

6x－（4＋x/60)*30|=45

Solution: x 30 or x = 25 11

Therefore, the time of the meeting may be 1 hour of Hengzen, or it may be 355 660 hours.