# How many times do a clock's hands overlap in a day?

# How many times do the hands of a clock overlap in a day?

A Better Approach (with reasoning) There are 2 cases depending on the working of the watch. Case 1: the movement of the second, minute and hour hands are continuous (not s…tep-wise or click-based) Answer is: the hour and minute hands overlap every hour. .
Case 2 (very unusual): the hour hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the minute hand crosses (or reaches 12) and the minute hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the second hand crosses (or reaches 12). Answer: every 65 minutes. Interviewers often expect this answer cos they do not think accurately. The exact times are: 0000 (12:00 AM) 0105 (01:05 AM) 0211 (02:11 AM) 0316 (03:16 AM) 0422 (04:22 AM) 0527 (05:27 AM) 0633 (06:33 AM) 0738 (07:38 AM) 0844 (08:44 AM) 0949 (09:49 AM) 1055 (10:55 AM) 1200 (12:00 PM) 1305 (01:05 PM) 1411 (02:11 PM) 1516 (03:16 PM) 1622 (04:22 PM) 1727 (05:27 PM) 1833 (06:33 PM) 1938 (07:38 PM) 2044 (08:44 PM) 2149 (09:49 PM) 2255 (10:55 PM) .
Reasoning for Case 1: ----------------------------- When do they overlap? At every (n + (n/11)) hours where n = 0, 1, 2, 3, ..., 24. How did I find this out? The following is the reasoning i used: At 0000, the hour and minute hands overlap. So number of overlaps now is 1. The minute hand races away and never again overlaps during the next one hour. Now, the minute hand moves at 360 o /hour and the hour hand moves at 30 o /hour. At 0100, the hour hand would be the 1 mark (or 30 o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0100), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as: 30 o + (30 o )x(T) = (360 o )x(T) How did I get to this equation? Note that, at 0100, when the minute hand starts moving from the 12 mark, the hour hand is already ahead of the minute hand by 30 o . If the minute hand moves at a speed of 360 o /hour, then in some time (T), it would cover (360 o )(T) degrees. If the hour hand hand moves at a speed of 30 o /hour, then in the same time (T), it would cover (30 o )(T) degrees. Since the hour hand is already 30 o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30 o + the number of degrees covered in time T) which is (30 o + (30 o )x(T)). Now the condition when the two hands will overlap is that they should have covered the same number of degrees at a moment (or) No. of degrees covered by minute hand = No. of degrees covered by hour hand (or) .
30 o + (30 o )x(T) = (360 o )x(T) .
If you solve this equation to find the value of T, you would get 30 o = (360 o )x(T) - (30 o )x(T) 30 o = (360 o - 30 o ) x T 30 o = 330 o x T (or) T = 30 o /330 o T = 1/11 At 0200, the hour hand would be the 2 mark (or 60 o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0200), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as: 60 o + (30 o )x(T) = (360 o )x(T) Solving this equation, you will the value of T = 2/11 At 0300, using the same reasoning (the hour hand at 90 o past the 12 mark) and modifying the equation accordingly (90 o + (30 o )x(T) = (360 o )x(T)), you would get the answer for T = 3/11. In general, the value for T for every hour is T = n/11 where n = 0, 1, 2, 3, ..., 24. So the exact time when the two hands overlap can be written as: the hour (n) + the time taken during that hour (T) .
(or) n + n/11 where n = 0, 1, 2, 3, ..., 24. .
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 22 is correct . The hands overlap about every 65 minutes, not every 60 minutes. In a day, the hands would only overlap 22 times, as illustrated in the table above. I would propose that the hands always overlap, as they're both attached at the center of the dial. If you didn't want to be facetious (or, at least, less facetious), you would still have to ask how many hands were on the clock. It may have a second hand, for example, or be digital (no hands at all).

# Imagine an analog clock set to 12 o'clock Note that the hour and minute hands overlap. How many times each day do both the hour and minute hands overlap How would you determine the exact times?

The hands do not overlap between 11 to 12, so only 22 overlaps per day. ---------------------------------.
1 of 2: Cjcarr2000 answer: --------------------------------- … It occurs 12 times a day Overlap Time = Hour : (Hour * 5) 12:00, 1:05, 2:10, 3:15, 4:20, etc. --------------------------------- 2 of 2: Stormnoone answer: --------------------------------- It occurs 22 times a day. (as Digbybare said, and not 24 as i was said at first) Exact times = Hour : ( Hour * 65 / 12 ) that is, when it is "1:05", the hour hand has moved from '1' to '2' (that total distance is 5 minutes) as the minute hand has moved from '12' to '1' (that is exactly 5 minutes, or Hour * 5); so the Hour hand must has moved "Hour / 12" of the 5 minutes: = Hour : [ (Hour * 5) + (Hour / 12) * 5 ] = Hour : [ (Hour * 5) + (Hour * 5) / 12 ] => we take out the common (Hour * 5) = Hour : [ (Hour * 5) * (1 + 1/12) ] = Hour : [ (Hour * 5) * 13/12 ] = Hour : (Hour * 65/12). ------------------------- So the Successive exact time should be 1 : ~5.4 2 : ~10.8 3 : 16.25 4 : ~21.67 5 : ~27.08 6 : 32.5 ......,and so on.

# What is the speed of a clock's second hand?

A clock's second hand makes one complete revolution each minute. Thus, by definition, it is rotating at one revolution per minute or one RPM. That's its "rotational velocity" …and it is the same no matter how big or small the clock might be. The actual velocity that the tip of the second hand might trace out as it revolves around the center of the clock will vary with the length of the second hand. The longer the hand, the faster the tip moves around the circumference.

# How many times a day does a clock's hands overlap?

Answer is: the hour and minute hands overlap every 65 minutes. Interviewers often expect this answer because they do not think accurately. The exact times are: 0000 (12:00… AM) 0105 (01:05 AM) 0211 (02:11 AM) 0316 (03:16 AM) 0422 (04:22 AM) 0527 (05:27 AM) 0633 (06:33 AM) 0738 (07:38 AM) 0844 (08:44 AM) 0949 (09:49 AM) 1055 (10:55 AM) 1200 (12:00 PM) 1305 (01:05 PM) 1411 (02:11 PM) 1516 (03:16 PM) 1622 (04:22 PM) 1727 (05:27 PM) 1833 (06:33 PM) 1938 (07:38 PM) 2044 (08:44 PM) 2149 (09:49 PM) 2255 (10:55 PM) .
Case 1: the movement of the second, minute and hour hands are continuous (not step-wise or click-based) .
Case 2 (very unusual): the hour hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the minute hand crosses (or reaches 12) and the minute hand jumps from 1 to 2, 2 to 3, ... and so on, as soon as the second hand crosses (or reaches 12). Reasoning for Case 1: ----------------------------- When do they overlap? At every (n + (n/11)) hours where n = 0, 1, 2, 3, ..., 24. How did I find this out? The following is the reasoning i used: At 0000, the hour and minute hands overlap. So number of overlaps now is 1. The minute hand races away and never again overlaps during the next one hour. Now, the minute hand moves at 360 o /hour and the hour hand moves at 30 o /hour. At 0100, the hour hand would be the 1 mark (or 30 o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0100), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as: 30 o + (30 o )x(T) = (360 o )x(T) How did I get to this equation? Note that, at 0100, when the minute hand starts moving from the 12 mark, the hour hand is already ahead of the minute hand by 30 o . If the minute hand moves at a speed of 360 o /hour, then in some time (T), it would cover (360 o )(T) degrees. If the hour hand hand moves at a speed of 30 o /hour, then in the same time (T), it would cover (30 o )(T) degrees. Since the hour hand is already 30 o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30 o + the number of degrees covered in time T) which is (30 o + (30 o )x(T)). Now the condition when the two hands will overlap is that they should have covered the same number of degrees at a moment (or) No. of degrees covered by minute hand = No. of degrees covered by hour hand (or) .
30 o + (30 o )x(T) = (360 o )x(T) If you solve this equation to find the value of T, you would get 30 o = (360 o )x(T) - (30 o )x(T) 30 o = (360 o - 30 o ) x T 30 o = 330 o x T (or) T = 30 o /330 o T = 1/11 At 0200, the hour hand would be the 2 mark (or 60 o from the 12 mark) and the minute hand would be at the 12 mark. Starting at this position (at 0200), they would overlap when the number of degrees moved by both the minute and the hour hand are the same. Let them overlap at some time, say T, then I can write them in equation form as: 60 o + (30 o )x(T) = (360 o )x(T) Solving this equation, you will the value of T = 2/11 At 0300, using the same reasoning (the hour hand at 90 o past the 12 mark) and modifying the equation accordingly (90 o + (30 o )x(T) = (360 o )x(T)), you would get the answer for T = 3/11. In general, the value for T for every hour is T = n/11 where n = 0, 1, 2, 3, ..., 24. So the exact time when the two hands overlap can be written as: the hour (n) + the time taken during that hour (T) .
(or) n + n/11 where n = 0, 1, 2, 3, ..., 24. .
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 22 is correct . I would propose that the hands always overlap, as they're both attached at the center of the dial. If you didn't want to be facetious (or, at least, less facetious), you would still have to ask how many hands were on the clock. It may have a second hand, for example, or be digital (no hands at all). 22 times a day if you only count the minute and hour hands overlapping. The approximate times are listed below. (For the precise times, see the related question.) 2 times a day if you only count when all three hands overlap. This occurs at midnight and noon. 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 A really simple way to see this is to imagine that the two hands are racing each other around a track. Every time the minute hand 'laps' the hour hand, we have the overlaps we want. So, we can say that the number of laps completed by the minute hand every T hours, Lm = T laps. Since there are 12hours in a full rotation of the hour hand, that hand only rotates Lh = T/12 laps. In order for the first 'lapping' to occur, the minute hand must do one more lap than the hour hand: Lm = Lh +1, so we get T = T/12 + 1 and that tells us that the first overlap happens after T = (12/11) hours. Similarly, the 2nd lapping will occur when Lm = Lh + 2. In general, the 'Nth' lapping will occur when Lm = Lh +N, which means every N*(12/11) hours (for N = 0,1,2,3...). In other words, it will happen approximately every 1hr5mins27secs, starting at 00:00. In 24hours, this occurs a total of 24/(12/11) = 22 times. ====== The same thing is explained below in a slightly less intuitive manner, using angular frequency: ====== So we are looking at two rotating hands. Ultimately, its just the angles we care about. Let Î¸H represent the angle of the hours hand and Î¸M represent the angle of the minutes hand. You could also introduce the seconds hand but that makes the problem more complicated. For now, lets assume the question only cares about the minute and hour hands. Initially we might think we are looking for: Î¸H=Î¸M But this doesn't take into account that if one hand has "gone around" a few times, its angle will be different from a hand in the same position that hasn't "gone around" the same number of times. So we have to modify our goal. Instead we let the angles differ by an integer multiple of 2Ï (360Â°). Let us call this arbitrary integer z. Now our condition is: Î¸M-Î¸H=2Ïz You could subtract the two angles in either order but the reason I chose to subtract hours from minutes is because it will result in positive integers which is just simpler. The minute hand goes around more times, thus its angle is bigger, thus this order of subtraction is positive. Now we have to find out how these angles depend upon the time. Let us call our time t and measure it in hours. I omit units for simplicity. The hour hand goes around a full rotation (2Ï) once every 12 hours. So: Î¸H=(2Ï/12) t For those more versed in mathematics, 2Ï/12 is the "angular frequency" for the hour hand (usually denoted by Ï). Similarly the minute hand goes around a full rotation (2Ï) once every hour. So: Î¸M=2Ï t Plugging back in: Î¸M-Î¸H=2Ïz 2Ï t - (2Ï/12) t = 2Ïz t - t/12 = z (11/12) t = z Now we are ready to solve. The two hands overlap at every solution of this equation, so we want to know the number of solutions of this equation. But remember, we want to know how many times this happens in a single day, so t cant be bigger than 24 (remember we are measuring t in hours), and technically no smaller than 0 (assuming we start our clock at 0 hours). Since t and z are proportional, each solution for z corresponds to exactly one solution for t, and accordingly exactly one solution of the equation. Also, remember than z must be an integer. So if we wanted all the times we would just let z go from 0 (when t=0) up and solve for t and stop as soon as we passed t=24. Then of course we'd have to convert that into hour and minute format. However, we only care about the number of times this happens. So we can notice that as t increases, z is just keeping track of how many times the two hands have overlapped. When z=0 we get the first time, when z=1 we get the second time, and so on. Since t and z are directly proportional, t increases with z, thus z represents the number of times the hands have overlapped up until time t minus 1 (and starting from t=0). Since we don't want t to go past 24, we plug in 24 and solve for z which will tell us how many times this event has occurred from t=0 to t=24 (one day). (11/12)*24 = z 22 = z So this happens 22 times in a day. Technically this has 23 solutions (0 through 22) but the last one is for t=24 which has begun the next day. If we don't count that solution we are left with 22.â If we want the second hand to overlap as well, we have to go a bit further. First we note that the second hand makes a full rotation once every minute, thus 60 times an hour. From this we have: Î¸S=(2Ï*60) t We want the second and hour hands to overlap AND the minute and second hands to overlap. Those conditions can be summarized as follows, where x and y are positive integers: Î¸S-Î¸M=2Ïx Î¸S-Î¸H=2Ïy Plugging in our functions of t for the Î¸'s and solving for t we are left with: t=x/59 t=12y/719 We want our integers x and y to produce the same time (making all hands overlap at that time). So we want to set the two equations equal. Simplifying, we get; x=708y/719 708 and 719 are coprime (719 is prime and 708 is decomposed into 2^2*3*59). In fact 708y and 719 are coprime except for when y is an integer multiple of 719. Thus 708y/719 can only be reduced when y=719k for some integer k. In this case we have: x=708k The first solution is when k=0. Then x=0 and t=0 corresponding to midnight. The next solution is k=1. Then x=708 and t=12 corresponding to noon. The next solution is k=2 but this corresponds to t=24 which is (midnight for) the next day and due to the direct proportionality of t and k, every k from here on up will produce t's higher than 24. In summary, all three hands only overlap twice a day: at noon and midnight. â All of this assumes that the hands sweep continuously. So the math is more(?) complicated for those with fake Rolex's (or any ticking handed clocks). Starting at 00:00 (midnight) with overlapping hands. At noon the hands have overlapped an additional 12 times and at midnight another additional 12 so including the starting and finishing midnight overlaps, 25 times . .

# What is the speed of a clock's minute hand?

it moves at one click every 60 seconds

# How many time clock's needles will be togher?

In how long a time? An hour, day, week, month etc.

# How many times will the hand on a clock overlap during a year?

8,760 (365 days) 24 * 365. 8,784 (366 days) 24 * 366..... I think, not sure. 8P

# How many times do the minute hand and the hour hand overlap?

Once per hour.

# How many times does the minute hand of a clock overlap with the hour hand from 10am to 12am?

23 times in all. Note that from 11:00 to 11:59 (am or pm) the hands can never overlap. Thus from 10am to 11:59am, the hands will overlap just once at around 10:54am. The hand…s will overlap again at exactly 12:00pm (noon). And from 12:01pm to 12:am (midnight) the hands will overlap another 11 times. The following times are the approximate overlapping times. .
10:54am .
12:00pm (noon) .
1:06pm .
2:11pm .
3:16pm .
4:21pm .
5:27pm .
6:32pm .
7:38pm .
8:43pm .
9:49pm .
10:54pm .
12:00am (midnight).

# How many times do not a clock's hands overlap in a day?

Infinitely many. Unless you consider the Planck time as the smallest, indivisible unit of time so that time is not a continuous variable.

Answered

In Uncategorized

# How many times does the minute hand and the hour hand overlap in the 24-hour clock in a day?

Twenty-four times of course !.

Answered

In Geometry

# How many times do a clock's three hands overlap in a day?

All 3 hands overlap 24 times a day.

Answered

In Math History

# How many times after 300 will the hands of a clock overlap?

Infinitely many. They will not stop overlapping -from time to time - as long as the clock keeps on working.

Answered

# How many times do a clock's hands overlap in 1 day?

22 times

Answered

# How many times does the hands of the clock overlap in a day?

Overlap happens once 12/11 hour. So 24Ã·12/11=22 Then overlap occurs 22 or 21 times a day.