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

# 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.

# How many times in a day do the hour hand and minute hand point in the same direction?

22 times in a 24 hour day. At midnight (the start of the day) the minute hand and hour hand are both pointing at 12. It will happen every 65 minutes 27.27 seconds. Here are th…e times: 12:00:00 midnight1:05:272:10:553:16:224:21:495:27:166:32:447:38:118:43:389:49:0510:54:33 That's 11 times in the morning, then at 12 noon, the cycle repeats, so 22 times.

# How many times a day does a person need to do hand washing?

A person does not "need" to wash their hands on a daily basis, but rather "should" wash their hands whenever circumstances dictate. These instances include: when your hands… are dirtybefore eating or touching food (like if you're helping cook or bake, for example)after using the bathroomafter blowing your nose or coughingafter touching pets or other animalsafter playing outsidebefore and after visiting a sick relative or friend So the conclusion is that, you need to wash your hands a lot. Probably every three to four hours or so.

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

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.

# What time between 2 and 3 o' clock do the minute hand and the hour hand overlap each other?

If the hands start off together at midnight, say, then the minute hand must catch up with the hour hand 11 times before they both end up on top of each other again at no…on. Hence between one and two o'clock the time when the hands are in the same place is 60/11 = 5.4545... minutes past the hour, between two and three o'clock 2*60/11 = 10.9090... minutes past the hour and so on. So the time you want is 10.91 minutes past 2 o'clock which, to the nearest second, is 2:10:54.

# What type of angle would be formed by a clock's hands when it is 3?

The hands form a right angle . At 3 o'clock, the minute hand points straight up, and the hour hand points straight to the right.

# How many times does the second hand move around the clock face in one day?

Once around in every minute, sixty minutes in an hour, and 24 hours in a day, so 1 X 60 X 24 = 1440.

# How many times does the second hand go around the clock a day?

Once per Minute -or- 60 times per hour 60x24 1440 Times a day (Also how many minutes that are in a day)

# How many times in a day the hands of a clock are straight?

the hands of a clock are straight 1)when they overlap & face the same direction & 2)when the overlap & face opposite directions this hapns evry 65 …min(approx)for each case =>in 1 day 24 hrs=> 24x60 min therefore each case hapns (24x60)/65 times each day................= 22(approx) therefore total # of times = 22+ 22=44

# How many times in a day are the hour hand and minute hand of a clock at right angles?

I just did some tallying in my head, with pencil and paper handy. In a 24 hour day, I count 44 times. The continuous movement of the hour hand brings about something ana…logous to the "sidereal day" problem. Most of the time, there are 2 occurances per hour of 90 degrees between the hands. But after 2 PM, for example, there is an occurance before 2:30, and the next occurance is 3 PM. There is then one occurance after 3 PM, the next one being after 4 PM. The same thing happens again at 8 and 9. So it seems that a total of 44 is probably right. Another way to 'visualize' that 'twice per hour' can't always work is to see that occurances of 90 degrees must be farther apart than 30 minutes, because both hands are advancing, not just the minute hand. For some hours, the first occurance will be late enough in the hour that the next occurance is in the next hour. --------------------------------------------------------------------------------------- Let me show you a mathematical approach. Common sense dictates that the minute hand moves at a faster rate of 5.5 degrees a minute (because the hour hand moves 0.5 degrees a min and the minute hand moves 6 degrees a minute). We start at 12 midnight. The hands are together. For subsequent 90 degree angles to occur, the minute hand must "overtake" the hour hand by 90 degrees, then 270 degrees, then 360 + 90 degrees, then 360 + 270 degrees, then 360 + 360 +90 degrees.. and so on. This can be re-expressed as: (1)90, 3(90), 5(90), 7(90), 9(90), 11(90)... n(90). The number of minutes this takes to happen can be expressed as (1)90/5.5, 3(90)/5.5, 5(90)/5.5, 7(90)/5.5, 9(90)/5.5, 11(90)/5.5... n(90)/5.5. In one day, there are 24 hr * 60 mins = 1440mins To find the maximum value of n, n(90)/5.5 = 1440 n = 88 but as seen from above, n must be an odd number (by pattern recognition and logic) hence n must be the next smallest odd number (87) counting 1,3,5,7,9,11......87, we see that the number of terms = (87-1)/2 +1 = 44. In other words, the minute hand "overtakes" the hour hand on 44 occasions in 24 hours in order to give a 90 degree angle. Therefore the answer to your question is 44.

# 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 co…ntinuous (not step-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 360o/hour and the hour hand moves at 30o/hour. At 0100, the hour hand would be the 1 mark (or 30o 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: 30o + (30o)x(T) = (360o)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 30o. If the minute hand moves at a speed of 360o/hour, then in some time (T), it would cover (360o)(T) degrees. If the hour hand hand moves at a speed of 30o/hour, then in the same time (T), it would cover (30o)(T) degrees. Since the hour hand is already 30o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30o + the number of degrees covered in time T) which is (30o + (30o)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) 30o + (30o)x(T) = (360o)x(T) If you solve this equation to find the value of T, you would get 30o = (360o)x(T) - (30o)x(T) 30o = (360o - 30o) x T 30o = 330o x T (or) T = 30o/330o T = 1/11 At 0200, the hour hand would be the 2 mark (or 60o 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: 60o + (30o)x(T) = (360o)x(T) Solving this equation, you will the value of T = 2/11 At 0300, using the same reasoning (the hour hand at 90o past the 12 mark) and modifying the equation accordingly (90o + (30o)x(T) = (360o)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).

# How many times a day is the hour hand opposite the minute hand?

22 times a day (11 times every 12 hours) Approximate times: 12:32:43 1:38:10 2:43:38 3:49:05 4:54:32 6:00:00 7:05:27 8:10:54 9:16:21 10:21:49 11:27:16 (see… the related questions below)

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# How many times does the average person wash their hands a day?

The correct answer would be 3.

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# 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 360o/hour and the hour hand moves at 30o/hour. At 0100, the hour hand would be the 1 mark (or 30o 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: 30o + (30o)x(T) = (360o)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 30o. If the minute hand moves at a speed of 360o/hour, then in some time (T), it would cover (360o)(T) degrees. If the hour hand hand moves at a speed of 30o/hour, then in the same time (T), it would cover (30o)(T) degrees. Since the hour hand is already 30o ahead from the 12 mark, the total degrees covered by the hour hand from the 12 mark would then be (30o + the number of degrees covered in time T) which is (30o + (30o)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) 30o + (30o)x(T) = (360o)x(T) If you solve this equation to find the value of T, you would get 30o = (360o)x(T) - (30o)x(T) 30o = (360o - 30o) x T 30o = 330o x T (or) T = 30o/330o T = 1/11 At 0200, the hour hand would be the 2 mark (or 60o 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: 60o + (30o)x(T) = (360o)x(T) Solving this equation, you will the value of T = 2/11 At 0300, using the same reasoning (the hour hand at 90o past the 12 mark) and modifying the equation accordingly (90o + (30o)x(T) = (360o)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 θH represent the angle of the hours hand and θM 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: θH=θM 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 2π (360°). Let us call this arbitrary integer z. Now our condition is: θM-θH=2πz 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 (2π) once every 12 hours. So: θH=(2π/12) t For those more versed in mathematics, 2π/12 is the "angular frequency" for the hour hand (usually denoted by ω). Similarly the minute hand goes around a full rotation (2π) once every hour. So: θM=2π t Plugging back in: θM-θH=2πz 2π t - (2π/12) t = 2πz 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.■ 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: θS=(2π*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: θS-θM=2πx θS-θH=2πy Plugging in our functions of t for the θ'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. ■ 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.

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# How many times in a day does a clock hands pass each other?

24. They pass each other after every hour.

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# How many times the hour hand go around the clock in 1 day?

twice i think

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In Computers

# 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 velo…city" 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.