Try to decide for yourself!
If something doesn’t work out, don’t despair, the answer and solution are located below.

1. How many times a day do clock readings have the property that by swapping the minute and hour hands we will arrive at a meaningful clock reading?

2. How many times a day do the hour and minute hands form a right angle?

3. How many minutes later will the (normal) clock hands overlap again after alignment?

4. How many times is the number showing how many times the speed of the second hand is greater than the speed of the minute hand, greater than the number showing how many times the speed of the minute hand is greater than the speed of the hour hand?

5. How many times will the hour hands be on top of each other in 12 hours?

6. Some work was started at the fifth hour and completed at the eighth hour, and the clock readings at the beginning and end of the work are converted into each other if the hour and minute hands are swapped. Determine the duration of the work and show that at the beginning and at the end of the work the arrows were equally deviated from the vertical direction.

7. How many times a day does the minute hand overtake the hour hand? What about a second?

8. The clock struck midnight. How many times and at what points in time before the next midnight will the hour and minute hands be aligned?

9. Between what numbers is the second hand located when the hour hand first aligns with the minute hand in the afternoon?

10. Why do the clock hands move from left to right (clockwise), and not vice versa?

11. On a watch with three hands - hour, minute and second - at 12 o'clock all three hands coincide. Are there other times when all three arrows coincide?

12. Problem proposed Lewis Carroll : Which clocks tell time more accurately: those that are behind by a minute per day, or those that do not go at all?

13. How many degrees does the minute hand rotate per minute? Hour hand?

14. Determine the angle between the hour and minute hands of a clock indicating 1 hour 10 minutes, provided that both hands move at constant speeds.

15.

16. But you probably noticed that this is not the only moment when the hands of the clocks meet: they overtake each other several times during the day. Can you point out all the times this happens?

17. When will the next meeting take place?

18. At 6 o'clock, on the contrary, both hands are directed in opposite directions. But does this only happen at 6 o’clock or are there other moments when the hands are positioned like this?

19. I looked at the clock and noticed that both hands were the same distance from the number 6, on both sides of it. What time was this?

20. At what time is the minute hand ahead of the hour hand by exactly the same amount as the hour hand is ahead of the number 12 on the dial? Or maybe there are several such moments a day or none at all?

21. What angle does the clock hand make at 12:20 o'clock?

22. Find the angle between the hour and minute hands a) at 9 o'clock 15 minutes; b) at 14:12?

23. When the angle between the hour and minute hands of a clock is greater than a) at 13:45 or 22:15; b) at 13:43 or 22:17; c) t minutes after noon or t minutes before midnight?

24. The clock hands have just aligned. After how many minutes will they “look” in opposite directions?

25. How can we explain that in a working watch the minute hand has passed 6 minutes in one second?

26. Using a precision chronometer, it was established that the hour and minute hands of a clock running evenly (but at the wrong speed!) coincide every 66 minutes. How many minutes per hour is this clock fast or slow?

27. In Italy they produce watches in which the hour hand makes one revolution per day, and the minute hand makes 24 revolutions per day, and, as usual, the minute hand is longer than the hour hand (in a regular watch, the hour hand makes two revolutions per day, and the minute hand makes 24). Let's consider all the positions of the two hands and the zero division, which are found on both Italian watches and ordinary ones. How many such provisions are there? (The zero mark marks 24 hours in Italian watches and 12 hours in regular watches).

28. Vasya measured with a protractor and wrote down in a notebook the angles between the hour and minute hands, first at 8:20, and then at 9:25. After that, Petya took his protractor. Help Vasya find the angles between the arrows at 10:30 and 11:35.

29. How many times do the minute and hour hands of a clock coincide from 12:00 to 23:59?

30. It's noon. When will the hour and minute hands coincide next time?

31. Indicate at least one point in time other than 6:00 and 18:00 when the hour and minute hands of a regularly running clock point in opposite directions.

32. When Petya began to solve this problem, he noticed that the hour and minute hands of his watch formed a right angle. While he was solving it, the angle was always obtuse, and the moment Petya finished solving it, the angle became right again. How long did Petya spend solving this problem?

33. Petya woke up at eight o'clock in the morning and noticed that the hour hand of his alarm clock bisected the angle between the minute hand and the bell hand pointing to the number 8. After what time should the alarm clock ring?

34. Kolya went for mushrooms between eight and nine o'clock in the morning at the moment when the hour and minute hands of his watch were aligned. He returned home between two and three o'clock in the afternoon, while the hands of his watch were directed in opposite directions. How long did Kolya’s walk last?

35. The student started solving the problem between 9 and 10 o'clock and finished between 12 and 13 o'clock. How long did it take him to solve the problem if during this time the hour and minute hands of the clock swapped places?

36. How many times during the day do the hour and minute hands of a properly running clock form an angle of 30 degrees?

37. There is a clock in front of you. How many hand positions are there that cannot tell the time unless you know which hand is the hour hand and which hand is the minute hand? (It is believed that the position of each of the arrows can be determined accurately, but it is impossible to monitor how the arrows move.)

38. In the world of the Antipodes, the minute hand of a clock moves at normal speed, but in the opposite direction. How many times per day the hands of the antipodean clocks a) coincide; b) opposite?

39. How many times a day can antipodean clocks be indistinguishable from normal ones (if you don’t know what time it really is)?

40. At noon, a fly sat on the second hand of the clock and drove off, adhering to the following rules: if it overtakes some hand or is overtaken by some hand (in addition to the second hand, the clock has hour and minute hands), then the fly crawls onto this hand. How many circles will a fly travel in an hour?

Pattern of time

Find out the pattern in the time change on the clock and determine what the clock at number five should show.

OGE tasks

Unified State Exam assignments

1. The clock with hands shows 8 hours 00 minutes. In how many minutes will the minute hand line up with the hour hand for the fourth time?

This task is no more difficult than the task of moving in a circle. Our hour and minute hands move in a circle. The minute hand travels a full circle in an hour, that is, 360°. Means, its speed is 360° per hour. The hour hand moves through an angle of 30° per hour (this is the angle between two adjacent numbers on the dial). Means, its speed is 30° per hour.

At 8:00 a.m. the distance between the hands is 240°:

Let the minute hand meet the hour hand for the first time after t hours. During this time, the minute hand will travel 360°t, and the hour hand 30°t, and the minute hand will travel 240° more than the hour hand. We get the equation:

360°t-30°t=240°

t=240°/330°=8/11

That is, after 8/11 hours the hands will meet for the first time.

Now, until the next meeting, the minute hand will travel 360° more than the hour hand. Let this happen in x hours.

We get the equation:

360°x-30°x=360°. Hence x=12/11. And so on two more times.

We get that the minute hand will align with the hour hand for the fourth time in 8/11+12/11+12/11+12/11= 4 hours= 240 minutes.

Answer: 240 min.

2. The clock with hands shows 1 hour 35 minutes. In how many minutes will the minute hand line up with the hour hand for the tenth time?

In this problem, we will express the speed of movement of the arrows in degrees/minute.

The speed of the minute hand is 360˚/60=6˚ per minute.

The speed of the hour hand is 30˚/60=0.5˚ per minute.

At 0 o'clock the position of the hour and minute hands coincided. 1 hour 35 minutes is 95 minutes. During this time, the minute hand moved 95x6=570˚=360˚+210˚, and the hour hand moved 95x0.5˚=47.5˚. And we have this picture:

The hands will meet for the first time after a time when the hour hand turns by , and the minute hand turns by 150˚+47.5˚ more. We get the equation for:

The next time the hands meet is when the minute hand passes one circle longer than the hour hand:

And so 9 times.

The minute hand will align with the hour hand for the tenth time in minutes

1. in 12 hours 132, in 24 hours 264 moments plus 22 overlays, total 286

2. The hour hand makes 2 revolutions per day, and the minute hand makes 24 revolutions. From here, the minute hand overtakes the hour hand 22 times and each time two right angles are formed with the hour hand, i.e. answer - 44 .

3. It is not difficult to figure out that this will happen after 1 hour 5 5/11 minutes, that is, at 2 hours 10 10/11 minutes. The next one is after another 1 hour 5 5/11 minutes, that is, at 3 hours 16 4/11 minutes, etc. All meetings, as you can easily see, will be 11; The 11th will occur 1 1/11 -12 hours after the first, that is, at 12 o'clock; in other words, it coincides with the first meeting, and further meetings will be repeated again at the same moments.

Here are all the moments of the meetings:

1st meeting - at 1 hour 5 5/11 minutes

2nd " - "2 hours 10 10/11 "

3rd " - "3 hours 16 4/11 "

4th " - "4 hours 21 9/11 "

5th " - "5 o'clock 27 3/11 "

6th " - "6 o'clock 32 8/11 "

2 hours 46, 153 min.

7. The hour hand makes 2 revolutions per day, and the minute hand makes 24 revolutions. From here the minute hand overtakes the hour hand 22 times.

9 . 4 and 5

10. This is exactly how the shadow moves in the very first hours - the sun. And then mechanical watches copied the direction of movement of the hands. By the way, in the Southern Hemisphere the opposite is true - the shadow in a sundial moves counterclockwise. In an hour, the minute hand makes a full revolution. This means that in a minute it rotates through 1/60th of an angle of 360°, that is, 6°. The hour hand travels 1/12 of the circle in an hour, that is, it moves 12 times slower than the minute hand. In a minute it rotates 0.5°.

14 . At 1:00 the minute hand was 30° behind the hour hand. In the 10 minutes that have passed since this moment, the hour hand will “travel” 5°, and the minute hand will “travel” 60°, so the angle between them is 60° – 30° – 5° = 25°.

15 . Let x be the period of time in minutes that must pass before the arrows are placed on the same straight line and directed in different directions. During this time, the minute hand will have time to travel x minute divisions of the dial, and the hour hand will have time to travel x/12 minute divisions. When the hands are placed on the same straight line and directed in different directions, they will be separated by 30 minute divisions of the dial. This means that at this time x – x/12 = 30, hence x = 32 (8/11). After 32 (8/11) minutes the arrows will “look” in opposite directions.

16 . Let's start watching the movement of the hands at 12 o'clock. At this moment, both arrows cover each other. Since the hour hand moves 12 times slower than the minute hand (it describes a full circle at 12 o’clock, and the minute hand at 1 hour), then, of course, the hands cannot meet during the next hour. But an hour passed; the hour hand is at number 1, having made 1/12 of a full revolution; The minute clock has made a full revolution and stands again at 12 - 1/12 of a circle behind the hour clock. Now the conditions of the competition are different than before: the hour hand moves slower than the minute hand, but it is ahead, and the minute hand must catch up with it. If the competition lasted a whole hour, then during this time the minute hand would go a full circle, and the hour hand would make 1/12 of a circle, that is, the minute hand would make 11/12 of a circle more. But in order to catch up with the hour hand, the minute hand needs to travel more than the hour hand, only by that 1/12th of a circle that separates them. This will take time not a whole hour, but less by the same amount of time as 1/12 is less than 11/12, that is, 11 times. This means that the hands will meet in 1/11 of an hour, that is, in 60/11 = 5 5/11 minutes. So, the meeting of the hands will occur 5 5/11 minutes after 1 hour has passed, that is, at 5 5/11 minutes of the second.

21. Answer: It is not difficult to figure out that this will happen after 1 hour 5 5/11 minutes, that is, at 2 hours 10 10/11 minutes. The next one is after another 1 hour 5 5/11 minutes, that is, at 3 hours 16 4/11 minutes, etc. All meetings, as you can easily see, will be 11; The 11th will occur 1 1/11 -12 hours after the first, that is, at 12 o'clock; in other words, it coincides with the first meeting, and further meetings will be repeated again at the same moments. Here are all the moments of the meetings:

24. Let both hands stand at 12, and then the hour hand moves away from 12 by a certain part of a full revolution, which we will denote by the letter x. During the same time, the minute hand managed to turn 12x. If no more than one hour has passed, then to satisfy the requirement of our task it is necessary that the minute hand is at a distance from the end of the whole circle by the same amount as the hour hand has time to move away from the beginning; in other words: 1 - 12 x = x Hence 1 = 13 x. Therefore, x = 1/13 of a whole revolution. The hour hand completes this fraction of a revolution at 12/13 o'clock, that is, it shows 55 5/13 minutes past midnight. The minute hand at the same time has traveled 12 times more, that is, 12/13 of a full revolution; both arrows, as you can see, are equally spaced from 12, and therefore equally spaced from 6 on opposite sides. We found one position of the arrows - exactly the one that occurs during the first hour. During the second hour, a similar situation will occur again; we will find it, reasoning according to the previous one, from the equality 1 - (12x - 1) = x, or 2 - 12x = x, whence 2 = 13x, and, therefore, x = 2/13 of a full revolution. In this position, the hands will be at 1 11/13 o'clock, that is, at 50 10/13 minutes past. The third time the hands will take the required position, when the hour hand moves away from 12 to 3/13 of a full circle, that is, 2 10/13 hours, etc. There are 11 positions, and after 6 o’clock the hands change places: the hour hand takes those places in which the minute hand was previously, and the minute hand takes the place of the hour hand. If you carefully watch the clock, then perhaps you have seen the exact opposite arrangement of the hands as described now: the hour hand is ahead of the minute hand by the same amount, by how much has the minute moved forward from the number 12. When does this happen? Answer: For the first time, the required arrangement of the hands will be at that moment, which is determined by the equality: 12x - 1 = x/2, whence 1 = 11 ½ x, or x = 2/23 of a whole revolution, that is, 1 1/23 hours after 12. This means that at 1 hour 21 4/23 minutes the hands will be positioned as required. Indeed, the minute hand should be in the middle between 12 and 1 1/23 o'clock, that is, at 12/23 o'clock, which is exactly 1/23 of a full revolution (the hour hand will travel 2/23 of a full revolution). The second time the arrows will be positioned in the required manner at the moment, which is determined from the equality: 12x - 2= x/2, from which 2 = 11 1/2 x and x = 4/23; the required moment is 2 hours 5 5/23 minutes. The third desired moment is 3 hours 7 19/23 minutes, etc.

Problem about clock hands. Task 11

1. Task 11 (No. 99600)

The clock with hands shows 8 hours 00 minutes. In how many minutes will the minute hand line up with the hour hand for the fourth time?

This task is no more difficult than the task of moving in a circle. Our hour and minute hands move in a circle. The minute hand travels a full circle in an hour, that is, 360°. Means, its speed is 360° per hour. The hour hand moves through an angle of 30° per hour (this is the angle between two adjacent numbers on the dial). Means, its speed is 30° per hour.

At 8:00 a.m. the distance between the hands is 240°:

Let the minute hand meet the hour hand for the first time after t hours. During this time, the minute hand will travel 360°t, and the hour hand 30°t, and the minute hand will travel 240° more than the hour hand. We get the equation:

360°t-30°t=240°

t=240°/330°=8/11

That is, after 8/11 hours the hands will meet for the first time.

Now, until the next meeting, the minute hand will travel 360° more than the hour hand. Let this happen in x hours.

We get the equation:

360°x-30°x=360°. Hence x=12/11. And so on two more times.

We get that the minute hand will align with the hour hand for the fourth time in 8/11+12/11+12/11+12/11= 4 hours= 240 minutes.

Answer: 240 min.

2. Task 11 (№ 114773). The clock with hands shows 1 hour 35 minutes. In how many minutes will the minute hand line up with the hour hand for the tenth time?

In this problem, we will express the speed of movement of the arrows in degrees/minute.

The speed of the minute hand is 360˚/60=6˚ per minute.

The speed of the hour hand is 30˚/60=0.5˚ per minute.

At 0 o'clock the position of the hour and minute hands coincided. 1 hour 35 minutes is 95 minutes. During this time, the minute hand moved 95x6=570˚=360˚+210˚, and the hour hand moved 95x0.5˚=47.5˚. And we have this picture:

The hands will meet for the first time after a time when the hour hand turns by , and the minute hand turns by 150˚+47.5˚ more. We get the equation for:

Number six

Most unwarned people in response to the question; For this task, draw one of the outlines: 6 or VI.

This shows that you can see a thing 100 thousand times and still not know it. The fact is that usually there is no number six on the dial (of men's watches), because a second is placed in its place.

Three o'clock

In 720 days. During this time, the second watch will fall behind by 720 minutes, that is, exactly 12 hours; the third clock will advance by the same amount. Then all three clocks will show the same as on January 1, that is, the correct time.

Two hours

The alarm clock goes off by 3 minutes during the hour compared to the wall clock. For 1 hour, that is, for 60 minutes, it goes away within 20 hours. But during these 20 hours, the alarm clock moved ahead by 20 minutes compared to the correct time. This means that the hands were set correctly 19 hours 20 minutes ago, that is, at 11 hours 40 minutes.

What time is it?

Between 3 and 6 hours 180 minutes. It is not difficult to figure out that the number of minutes remaining until 6 hours will be found if 180-50, that is, 130, is divided into two parts, one of which is four times larger than the other. This means we need to find a fifth of 130. So, it was 26 minutes to six.

Indeed, 50 minutes ago there were 26 + 50 - 76 minutes left until 6 hours, and, therefore, after 3 hours 180-76 = 104 minutes had passed; this is four times the number of minutes now remaining until six.

When do the arrows meet?

Let's start watching the movement of the hands at 12 o'clock. At this moment, both arrows cover each other. Since the hour hand moves 12 times slower than the minute hand (it describes a full circle at 12 o’clock, and the minute hand at 1 hour), then, of course, the hands cannot meet during the next hour. But an hour passed; the hour hand is at number 1, having made 1/12 of a full revolution; The minute clock has made a full revolution and stands again at 12 - 1/12 of a circle behind the hour clock. Now the conditions of the competition are different than before: the hour hand moves slower than the minute hand, but it is ahead, and the minute hand must catch up with it. If the competition lasted a whole hour, then during this time the minute hand would go a full circle, and the hour hand would make 1/12 of a circle, that is, the minute hand would make 11/12 of a circle more. But in order to catch up with the hour hand, the minute hand needs to travel more than the hour hand, only by that 1/12th of a circle that separates them. This will take time not a whole hour, but less by the same amount of time as 1/12 is less than 11/12, that is, 11 times. This means that the hands will meet in 1/11 of an hour, that is, in 60/11 = 5 5/11 minutes.

So, the meeting of the hands will occur 5 5/11 minutes after 1 hour has passed, that is, at 5 5/11 minutes of the second.

When will the next meeting take place?

It is not difficult to figure out that this will happen after 1 hour 5 5/11 minutes, that is, at 2 hours 10 10/11 minutes. The next one - after another 1 hour 5 5/11 minutes, that is, at 3 hours 16 4/11 minutes, etc. All meetings, as you can easily see, will be 11; The 11th will occur 1 1/11 -12 hours after the first, that is, at 12 o'clock; in other words, it coincides with the first meeting, and further meetings will be repeated again at the same moments.

Here are all the moments of the meetings:

1st meeting - at 1 hour 5 5/11 minutes

2nd" - "2 hours 10 10/11"

3rd" - "3 hours 16 4/11"

4th" - "4 o'clock 21 9/11"

5th" - "5 o'clock 27 3/11"

6th" - "6 o'clock 32 8/11"

7th" - "7 o'clock 38 2/11"

8th" - "8 o'clock 43 7/11"

9th" - "9 o'clock 49 1/11"

10th" - "10 o'clock54 6/11"

11th "- "12 o'clock

When are the arrows pointing apart?

This problem is solved very similarly to the previous one. Let's start again at 12 o'clock, when both hands coincide. You need to calculate how long it will take for the minute hand to overtake the hour hand by exactly half a circle - then both hands will be directed in exactly opposite directions. We already know (see the previous problem) that over the course of a whole hour, the minute hand overtakes the hour hand by 11/12 of a full circle; to overtake her by just 1/2 of a lap, it will take less time than a whole hour - less by as many times as 1/2 is less than 11/12, that is, it will take only 6/11 hours. This means that after 12 o’clock the hands are positioned opposite each other for the first time after 6/11 hours, or 32 8/11 minutes. Look at the clock at 32 8/11 minutes past one and you will see that the hands are pointing in opposite directions.

Is this the only time when the arrows are positioned like this? Of course not. The hands take this position 32 8/11 minutes after each meeting. And we already know that there are 11 meetings in 12 hours; This means that the hands are also positioned apart 11 times within 12 hours. It is not difficult to find these moments:

12 o'clock + 32 8/11 min. = 12 hours. 32 8/11 min.

1 hour 5 5/11 min. + 32 8/11 min. = 1 hour 38 2/11 min.

2 hours 10 10/11 min. + 32 8/11 min. = 2 hours 43 7/11 min.

3 hours 16 4/11 min. + 32 8/11 min. = 3 hours 49 1/11 minutes, etc.

I’ll leave it to you to calculate the remaining moments yourself.

On both sides of six

This problem is solved in the same way as the previous one. Let's imagine that both hands stood at 12, and then the hour clock moved away from 12 by some part of a full revolution, which we will denote by the letter x. During the same time, the minute hand managed to rotate 12 x;. If no more than one hour has passed, then to satisfy the requirement of our task it is necessary that the minute hand is at a distance from the end of the whole circle by the same amount as the hour hand has time to move away from the beginning; in other words:

1 - 12 x = x

Hence 1 = 13 x (because 13 x - 12 x - x). Therefore, x = 1/13 of a whole revolution. The hour hand completes this fraction of a revolution at 12/13 o'clock, that is, it shows 55 5/13 minutes past midnight. The minute hand at the same time has traveled 12 times more, that is, 12/13 of a full revolution; both arrows, as you can see, are equally spaced from 12, and therefore equally spaced from 6 on opposite sides.

We found one position of the arrows - exactly the one that occurs during the first hour. During the second hour, a similar situation will occur again; we will find it, reasoning according to the previous one, from the equality

1 - (12x - 1) = x, or 2- 12x = x,

whence 2 = 13x (because 13x - 12x = x), and therefore x = 2/13 of a full revolution. In this position, the hands will be at 1 11/13 o'clock, that is, at 50 10/13 minutes past.

The third time the hands will take the required position, when the hour hand moves away from 12 to 3/13 of a full circle, that is, 2 10/13 hours, etc. There are 11 positions, and after 6 o’clock the hands change places: the hour hand takes those the places where the minute one was before, and the minute one takes the place of the hour one.

What time?

If we start watching the hands at exactly 12 o'clock, then during the first hour we will not notice the desired location. Why? Because the hour hand travels 1/12 of what the minute hand travels, and therefore lags behind it much more than is required for the desired location. Whatever angle the minute hand moves away from the 12, the hour hand will turn by 1/12 of this angle, and not by 1/2, as we require. But an hour passed; Now the minute hand is at 12, the hour hand at 1, 1/12 of a full turn ahead of the minute hand. Let's see if this arrangement of hands can occur during the second hour. Let us assume that this moment occurred when the hour hand moved away from the number 12 by a fraction of a revolution, which we denote by x. The minute hand managed to travel 12 times further in the same time, that is, 12x. If you subtract one full revolution from here, then the remainder 12x - 1 must be twice as large as x, that is, equal to 2x. We see, therefore, that 12x - 1 = 2x, from which it follows that one whole revolution is equal to 10x (indeed, 12x -10x = 2x).

But if 10x is equal to a whole revolution, then 1x = 1/10 of a revolution. Here is the solution to the problem: the hour hand has moved away from the number 12 by 1/10 of a full revolution, which requires 12/10 of an hour or 1 hour. 12 minutes. The minute hand will be twice as far from 12, that is, at a distance of 1/5 of a turn; this answers 60/5 = 12 minutes, as it should be.

We have found one solution to the problem. But there are others: the hands are positioned in the same way not just once, but several times over the course of 12 o’clock. Let's try to find other solutions.

To do this, we will wait 2 hours; the minute hand is at 12, and the hour hand is at 2. Reasoning from the previous one, we get the equality:

whence two whole revolutions are equal to 10x, and, therefore, x = 1/5 of a whole revolution. This corresponds to the moment 12/5 = 2 hours 24 minutes.

You can easily calculate the further points yourself. Then you will find that the arrows are positioned according to the requirement of the problem at the following 10 points:

at 1 hour 12 minutes at 7 hours 12 minutes

« 2 hours 24 « « 8 « 24 «

"3"36"" 9"36"

"4"48"" 10"48"

"6 o'clock"12"

The answers: “at 6 o’clock” and “at 12 o’clock” may seem incorrect, but only at first glance. Indeed: at 6 o’clock the hour hand is at 6, and the minute hand is at 12, that is, exactly twice as far away. At 12 o’clock, the hour hand is removed from 12 “by zero”, and the minute hand, if you like, by “two zeros” (because a double zero is the same as a zero); This means that this case, in essence, satisfies the conditions of the problem.

Vice versa

After the previous explanations, solving this problem is no longer difficult. It is easy to understand, reasoning as before, that for the first time the required arrangement of arrows will be at that moment, which is determined by the equality:

whence 1 = 11 ½ x, or x = 2/23 of a whole revolution, that is, 1 1/23 hours after 12. This means that at 1 hour 21 4/23 minutes the hands will be positioned in the required manner. Indeed, the minute hand should be in the middle between 12 and 1 1/23 o'clock, that is, at 12/23 o'clock, which is exactly 1/23 of a full revolution (the hour hand will travel 2/23 of a full revolution). The second time the arrows will be positioned in the required manner at the moment, which is determined from the equality:

whence 2 = 11 1/2 x and x = 4/23; the desired moment is 2 hours 5 5/23 minutes.

The third required moment is 3 hours 7 19/23 minutes, etc.

Three and seven

Usually the answer is: “7 seconds.” But this answer, as we will now see, is incorrect.

When the clock strikes three, we observe two intervals:

1) between the first and second blow;

2) between the second and third blow.

Both intervals last 3 seconds; This means that each lasts half as long - exactly 1 1/2 seconds.

When the clock strikes seven, there are six such intervals. Six times 1 1/2 seconds equals 9 seconds. Therefore, the clock “strikes seven” (that is, makes seven strokes) at 9 seconds.

The clock is ticking

Mysterious interruptions in the ticking of the clock are simply due to hearing fatigue. Our hearing, getting tired, becomes dull for a few seconds - and during these intervals we do not hear ticking. After a short time, fatigue passes and the former sensitivity is restored - then we again hear the clock ticking. Then fatigue sets in again, etc.

Number six Ask someone older you know how long they have owned a pocket watch. Suppose it turns out that he has had the watch for 15 years. Then continue the conversation something like this: - How many times a day do you look at your watch? . “Probably 20 times or so,” comes the answer. -...

Let us turn again to school tasks and intelligence tasks. One of these tasks is to find out what angle the minute and hour hands form between themselves on a mechanical watch at 16 hours 38 minutes, or one of the variations is to find out how much time it will be after the beginning of the first day when the hour and minute hands form an angle of 70 degrees.

The simplest question to which many people manage to give the wrong answer. What is the angle between the hour and minute hands on a clock at 15:15?

The answer zero degrees is not the correct answer :)

Let's figure it out.

In 60 minutes, the minute hand makes a full revolution around the dial, that is, it rotates 360 degrees. During the same time (60 minutes), the hour hand will travel only one-twelfth of the circle, that is, it will move by 360/12 = 30 degrees

As for the minute, everything is very simple. Compiling proportion minutes are related to the angle traversed as a complete revolution (60 minutes) is to 360 degrees.

Thus, the angle traveled by the minute hand will be minutes/60*360 = minutes*6

As a result, the conclusion Each minute passed moves the minute hand 6 degrees

Great! Now what about the sentry. But the principle is the same, only you need to reduce the time (hours and minutes) to fractions of an hour.

For example, 2 hours 30 minutes is 2.5 hours (2 hours and half), 8 hours and 15 minutes is 8.25 (8 hours and one quarter of an hour), 11 hours 45 minutes is 11 hours and three quarters of an hour, that is, 8.75)

Thus, the angle traversed by the clock hand will be hours (in fractions of an hour) * 360.12 = hours * 30

And as a consequence the conclusion Every hour passed moves the hour hand 30 degrees

angle between hands = (hour+(minutes /60))*30 -minutes*6

Where hour+(minutes /60)- this is the clockwise position

Thus, the answer to the problem: what angle will the hands make when the clock shows 15 hours 15 minutes, will be as follows:

15 hours 15 minutes is equivalent to the position of the hands at 3 hours and 15 minutes and thus the angle will be (3+15/60)*30-15*6=7.5 degrees

Determine the time by the angle between the arrows

This task is more difficult, since we will solve it in a general form, that is, determine all pairs (hour and minute) when they form a given angle.

So, let's remember. If time is expressed as HH:MM (hour:minute) then the angle between the hands is expressed by the formula

Now, if we denote the angle by the letter U and convert everything into an alternative form, we get the following formula

Or, getting rid of the denominator, we get the basic formula relating the angle between two hands and the positions of these hands on the dial.

note that the angle can also be negative, i.e. oh, within an hour we can meet the same angle twice, for example, an angle of 7.5 degrees can be at 15 hours 15 minutes and 15 hours and 17.72727272 minutes

If, as in the first problem, we were given an angle, then we get an equation with two variables. In principle, it cannot be solved unless one accepts the condition that the hour and minute can only be integers.

Under this condition we obtain the classical Diophantine equation. The solution to which is very simple. We will not consider them for now, but will immediately present the final formulas

where k is an arbitrary integer.

We naturally take the result of hours modulo 24, and the result of minutes modulo 60

Let's count all the options when the hour and minute hands coincide? That is, when the angle between them is 0 degrees.

At a minimum, we know two such points: 0 hours and 0 minutes and 12 noon 0 minutes. What about the rest??

Let's create a table showing the positions of the arrows when the angle between them is zero degrees

Oops! on the third line we have an error at 10 o'clock, the hands do not coincide. This can be seen by looking at the dial. What's the matter?? It seems like everything was calculated correctly.

But the whole point is that in the interval between 10 and 11 o’clock, in order for the minute and hour hands to coincide, the minute hand must be somewhere in the fractional part of a minute.

This can be easily checked using the formula by substituting the number zero instead of the angle, and the number 10 instead of the hour

we get that the minute hand will be located between (!!) divisions 54 and 55 (exactly at the position 54.545454 minutes).

That's why our latest formulas didn't work, since we assumed that hours and minutes are integers(!).

Problems that appear on the Unified State Exam

We will look at problems for which solutions are available on the Internet, but we will take a different route. Perhaps this will make it easier for that part of schoolchildren who are looking for a simple and easy way to solve problems.

After all, the more different options for solving problems, the better.

So, we know only one formula and we will only use it.

The clock with hands shows 1 hour 35 minutes. In how many minutes will the minute hand line up with the hour hand for the tenth time?

The reasoning of the “solvers” on other Internet resources made me a little tired and confused. For those “tired” like me, we solve this problem differently.

Let’s determine when in the first (1) hour the minute and hour hands coincide (angle 0 degrees)? We substitute the known numbers into the equation and get

that is, 1 hour and almost 5.5 minutes. is it earlier than 1 hour 35 minutes? Yes! Great, then we don’t take this hour into account in further calculations.

We need to find the 10th coincidence of the minute and hour hands, we begin to analyze:

for the first time the hour hand will be at 2 o'clock and how many minutes,

the second time at 3 o'clock and how many minutes

for the eighth time at 9 o'clock and for some minutes

for the ninth time at 10 o'clock and how many minutes

for the ninth time at 11 o'clock and for some minutes

Now all that remains is to find where the minute hand will be at 11 o'clock, so that the hands coincide

And now we multiply 10 times the revolution (which is every hour) by 60 (converting to minutes) and we get 600 minutes. and calculate the difference between 60 minutes and 35 minutes (which were specified)

The final answer was 625 minutes.

Q.E.D. There is no need for any equations, proportions, or which of the arrows moved at what speed. It's all tinsel. It is enough to know one formula.

A more interesting and complex task sounds like this. At 8 pm, the angle between the hour and minute hands is 31 degrees. How long will the hand show the time after the minute and hour hands form a right angle 5 times?

So in our formula, two of the three parameters are again known: 8 and 31 degrees. We determine the minute hand using the formula and get 38 minutes.

When is the nearest time when the arrows will form a right (90 degrees) angle?

That is, at 8 hours 27.27272727 minutes this is the first right angle in this hour and at 8 hours and 60 minutes this is the second right angle in this hour.

The first right angle has already passed relative to the given time, so we do not count it.

The first 90 degrees at 8 hours 60 minutes (we can say that exactly at 9-00) - once

at 9 o'clock and how many minutes - that's two

at 10 o'clock and how many minutes is it three

again at 10 and how many minutes is 4, so there are two coincidences at 10 o’clock

and at 11 o'clock and how many minutes is five.

It’s even easier if we use a bot. Enter 90 degrees and get the following table

that is, at 11 hours 10.90 minutes there will be just the fifth time when a right angle is again formed between the hour and minute hands.

We hope this analysis will help you both formulate tasks for students and easily solve similar tests for intelligence in the Unified State Exam.

Good luck with your calculations!