**Question:** How many times a day do the minute and hour hands of a clock overlap?

**Answer: **Did you think the answer was 24 times? Well if you did, it’s time you think again. Let’s do some math.

In T hours, the minute hand completes T laps. In the same amount of time, the hour hand completes T/12 laps.

The first time the minute and hour hands overlap, the minute hand would have completed 1 lap more than the hour hand. So we have T = T/12 + 1. This implies that the first overlap happens after T = 12/11 hours (~1:05 am). Similarly, the second time they overlap, the minute hand would have completed two more laps than the hour hand. So for N overlaps, we have T = T/12 + N.

Since we have 24 hours in a day, we can solve the above equation for N

24 = 24/12 + N

24 = 2 + N

**N = 22**

Thus, the hands of a clock overlap 22 times a day. Thus the hands of the clock overlap at 12:00, ~1:05, ~2:10, ~3:15, ~4:20, ~5:25, ~6:30, ~7:35, ~8:40, ~9:45, ~10:50. Note that there is no ~11:55. This becomes 12:00.

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The answer is 22, but for different reasons than stated above. The angular speed of the minute hand is 360°/hr, whereas the angular speed of the hour hand is 30°/hr. The hands overlap when the difference in the angles swept by the hands is a multiple of 360°, and the angles swept over an amount of time by the hands are given by the formula angle swept = time * angular speed. In other words, 360°/hr*t – 30°/hr*t = n360°, where t is time in hours and n is some integer. You can simplify this to 330°/hr*t = n360°, and then n=11t/12hr. There are 24 hours in a day, so n = 22, which is the total number of times the hands overlap. Therefore, the hands do NOT overlap at 1:05 (30° from 12), 2:10 (60° from 12), etc, they overlap at the t’s that correspond to 30° + 30° * 1/11 (slightly after 1:05), 60° + 30°*2/11 (slightly after 2:10), etc. The last overlap would be at angle of 330° + 30° *11/11, which = 360°, or a full cycle. Thus, 11 overlaps in 12 hours, 22 overlaps in 24 hours.

Thanks for the excellent post, I was looking for information similar to this, going to look at the various other articles.

the answer is 2

because at only 12 am and 12 pm,the condition is true..

otherwise in case of 1:05 when the minute hand is at 1,the hour hand is not at 1 but is slightly ahead of 1 as for every 1 degree rotation of minute hand,the hour hand rotates 1/12 degrees..

so practically it is never possible for both minute hand and hour hand to overlap each other completely exept at 12:00 am and pm.

The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e., at 12 o’clock).

AM

12:00

1:05

2:11

3:16

4:22

5:27

6:33

7:38

8:44

9:49

10:55

PM

12:00

1:05

2:11

3:16

4:22

5:27

6:33

7:38

8:44

9:49

10:55

The hands overlap about every 65 minutes, not every 60 minutes.

The hands coincide 22 times in a day.

I see how important it is to wear a watch at an interview lol.

@John It is not 23. You have think of it as 2 sets of 12 hours. If it is 12 hours, you logic is fine. You will get 12 – 1 (since it will 12am and 12pm will be the same). Now if you extend it to 2 set of 12 hours. Then you get 11*2=22. If this doesn’t convince you, try writing down all the matchs.

The error in the answer of 23 is that you are accounting for 12 O clock three times in the day, instead of just twice. In a true day, it should only be accounted for in the very first overlap at 12am. The next time 12am (not 12pm) occurs, it is the next day, and thus does not count.

22 is the correct answer, not 23. The time when t=0 has already been included in the 22 times. Look closer.

Oh, as an aside, the t=0 one is the correct one to count because technically 12:00am belongs to the day that follows it, not the day that precedes it. Hence the term ante-meridian, or “before the middle”. Likewise, noon is post-meridian and belongs to the second half of the day. Therefore, t=0 should be counted as the first overlap, and the following midnight (or t=24) should not be counted as belonging to this day.

I agree with abc, since at least the initial or the final overlap at t=24 or t = 0 has to be considered in the day, so its 23.

ABC is right!!!

uh each time the minute hand goes around the hour hand moves to a new spot, so they overlap once a minute,60 times and hour, and 1440 times in a day

As it is to be counted from 0 to 24 hours, it starts at 12 o clock and ends at 12 o clock. So its 23 times.

The site does not display my formula correctly~~ try it again:

24 * 60 t t ~= 22

anyway, i agree it is 22

After the previous overlap, if minute hand catches up hour hand again, it will move (1+1/12)*60 unit. So in 24 hours, the number of overlap will be calculated:

24 * 60 t t ~= 22

The essence of the problem is contained in the word “overlap”. As the word implies, it happens every time the minute hand laps the hour in their race around the clock. Since the minute hand completes 24 trips in the time that the hour hand completes only two, the minute hand laps the hour hand 22 times.

If we are counting the midnight as the second day then shouldn’t we count the midnight from previous day as today. In the end we count one of midnight (either the previous night or this night). So it ends up being 22.

Surely its 21, since the 22nd will actually be midnight of the 2nd day?

Excellent analysis!