“I bought this guide a few days ago to prepare for my interview with Oracle. Many of the questions they asked me were from this guide. I found this book absolutely great!”

The clock hands don’t overlap at 1:05, 2:10, 3:15 etc. Because, by the time the minutes hand reaches where the hours hand was, the hours hand has moved on, as it doesn’t remain fixed through the entire hour. the answer can be worked out in this way:
In every hour, as the hour hand moves across 5 minute spaces, the minute hand moves across 60. That means the minute hand gains 55 min sp. over hour hand in 60 min. Thus, between the nth and (n+1)st hr, to overlap the minute hand must gain the 5n min sp. that exists between them when the clock strikes n hrs. Apply unitary method and you get the time in which it does so i.e. 60n/11 min. So, the clock hands overlap at n hrs 60n/11 min. Obtain the values by putting m = 0 to 11, and you will find that from 00:00:00 to 23:59:59 (please note that 23:59:60 is equivalent to 24:00:00 or 00:00:00), the hands overlap 22 times.

Considering only minute and hour hand- during the course of each hour, minut hand overlaps the hour hand once and only once (except during the hour hand moves from 12:00 to 1:00). This maked only 24-1=23 overlaps in 24hours.

Submitted By: Debadyuti Banerjee — October 30, 2006

In any circular motion, if the ratio between 2 fellow is n:1 then for each lap of the slower fellow, faster fellow overlaps him n times, except for the last lap of the slower fellow, where the number of crossings are n-1. (The race just finishes before the n th cross.
slower fellow : hour hand (short)
faster fellow : minuite hand (long)
n = 12
Total laps for hour hand = 2
total overlaps = 12+11 = 23

The amount of times that the second hand overlaps each of the hour and the minute hand is twice a minute, i.e. two overlappings a minute (or 120 times an hour, or 2880 times a day). However, there are times where the minute hand overlaps the hour hand, and therefore the second hand overlaps them at the same time, taking away one overlapping every time this happens. We subtract one from the second hand overlappings every time the hour and minute hands overlap. Therefore, they cancel eachother out, and we are left with 2880 overlappings every day.

i donno how u can come to 23 or 22 hrs conclusion. all the math is not reqd. from 12 a.m to 12 p.m. how many times the hnds overlap. 12 (initial point), 1:06 ish, 2:14 ish..11:58 ish = 12. in a day it is 2 cycles i.e 24 duh!

Let say the speed of hour hand is 1x per hr, then the speed of minute hand is 12x per hr. That means the minute hand runs faster than hour hand 11x (12x - 1x) per hour. Start from 0:00AM (first over lap), time it needs for the minute hand to over lap again is: 12x/11x or 12/11 hr. So in 24 hr the number of overlap will be 1 + 24/12/11 = 23 times.

Assuming that hour hand, minute hand and seconds hand are of the same thickness. On a 24 hr cycle:

second hands actually move:
60 * 60 * 24 = 86400

since every 60 moves of the second hand, the minute hand only get to move once
Minute hand will have:
60* 24 = 1440;

Now the tricky part is the hour hand. It is not right to assume that hour clock only moves 24 times. Since in between each hour(example 1:00 to 2:00) hour hand actually moves the question is WHEN HE MOVES and HOW MANY movement he does before reaching the next hour. We can used this analogy below. For example 1:00 to 2:00:

distance, using the minutes hand from 1 to 2 is actually 5 minutes( or 5 ticks), using this we can also say that the hour hand will take 5 movements going to 1 to 2. In 1 hour the hour hand moves 5 times.

OK so I sat here for 24 hours and counted them.
Seriously I like the analysis of answer 5. For each hour except for the first and last in each 12 hours the hour and minute hands have to overlap exactly once. A day is 24 hours starting at midnight and ending at midnight. At those two times the hour and minute (and second hand if there is one) overlap. So those two overlaps cover the hours from midnight to 1 and from 11 to midnight when it does not otherwise overlap. However between 11 and noon and noon to 1 we have just one overlap at exactly noon since it is in the middle of the day you can’t count it twice as if the morning and afternoon were distinct unrelated entities when we are counting the overlaps in “a day.” We have 3 overlaps so far. For the other 20 hours, from 1 to 2, etc. till 10 to 11 the hour and minute hand will overlap exactly once. (The minute hand goes from 12 to 12 while the hour hand is a little past 1 or a little before 11 at the two ends and between those points for all the other hours so they have to overlap once.) So we have 23 overlaps in 24 hours counting the two midnights. If you were counting this over several consecutive days you could not count the midnight between two days twice, once for each day so it would average to 22 times. Problem says “a day” but if you interpret that to mean “each day.”
If there is a second hand then the result might be similar and depends mostly on the thickness of the hands. Every minute the second hand overlaps the minute hand once or twice or more if the minute hand is thick enough and the second hand moves in discreet jerky increments or if the hour minute hand overlap lasted longer than a minute. But if the hour and minute hands were too thin there is a chance that they would not overlap at the moment the second hand passed over the minute hand for some minutes and some of the hourly hour-minute hand overlaps would not get counted. Remember the second hand will overlap the minute hand every minute or so but we only care about when it overlaps the minute hand at the same time it is overlapping the hour hand and that only happens at 12 and every hour for the other 20 hours so assuming hour-minute overlap holds for exactly one minute the result is the same 23 overlaps.

I guess I should add as regards second hands the little clock that Google or Vista gives me has a thin second hand that moves a jerky second at a time and the hour and minute hands move smoothly and are a second (or minute) wide. The second hand can be considered to overlap the hour or minute hand once when it lands in the center of the hand or twice (in one rotation) at the edges of the hand. It will always overlap the minute hand at least once as it passes over.

If we consider the thickness of the hands to be very thin, and we are having 3 hands, i.e. Hour, Minute and Second. Now:

The Hour hand and Minute hand would be meeting exactly 11 times in 12 hours (Hour hand would have taken 1 clockwise round and Minute hand would have taken 12 clockwise rounds, so 12 - 1 = 11 rounds).

result: First time hour and minute hands overlap will be 12 Hours / 11 = 01:05:27.27. So at this time only hour and minute hands would be overlapping and second hand will not be any near to them. Similarly for 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th and 10th overlap of hour and minute hand the Second hand wont be any nearby. So all 3 hands (hour, minute and Second) overlap only 2 times i.e. (0:0:0 and 12:0:0).

Also we all know when we get our watches repaired, normally the repairman overlaps all the three hands to 12.

If we are considering that the second hand is not present, then the rest two overlaps 22 times in 24 hours.

There again is a catch, if we check the angles by which the hour hand and minute hand moves.

The second hand moves 6 degree in a second. In that time the minute hand will move 6/60 degrees. and the hour hand will move 6/(60*12) degrees. now taking these things in the considerations. if we check the positions of the hour and minute hand in terms of angle from the marker 12, for our first rendezvous time, i.e. 01:05:27.27 sec.
first thing that comes to my mind is that, there is fraction in the seconds. So that time can’t be measured. there will be no exact overlap. now lets calculate the angles:

1 hour 5 mins and 27 seconds = 3600 + 5*60 + 27 = 3927 seconds.

angle of hour hand = 3927 * 6/(60*12) = 32.725 degree.
angle of minute hand = 3927 * 6/60 = 392.7 degree
subtracting 360 degree from it we get - 32.7 degree.

So at 01:05:27 both hands don’t overlap. Now for 01:05:28 :
Angles : hour hand - 32.73333
minute hand - 32.8
so obviously they dont meet at 01:05:28 either.

So they overlap at 12:00 and 24:00 only. So the answer is 2 only.

The obvious guess is to write down a list of the places the hands will overlap and count them up: 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, 7:35, 8:40, 9:45, 10:50, 11:55, twice around gives 24 crossings.

Unfortunately this is the wrong answer, as the hour hand must move a bit more than one hour to catch up to the minute hand. Specifically it moves 12/11 of an hour or 65.454545… minutes. Thus the correct times are:

0:00:00 (midnight)
1:05:27
2:10:55
3:16:22
4:21:49
5:27:16
6:32:44
7:38:11
8:43:38
9:49:05
10:54:33
12:00:00 (noon)
13:05:27
14:10:55
15:16:22
16:21:49
17:27:16
18:32:44
19:38:11
20:43:38
21:49:05
22:54:33
24:00:00 (which doesn’t count because it’s in the next day)

So there are 22 crossings. Note that there is NO overlap in the expected area of 11:55 because it happens at 12:00. You can spin the time setting knob on an analog clock and see this easily.

Let “S” denote the arc which the small hand travels in one second:
S = 2*pi / (60*60*12) (1)
Let “B” denote the arc which the big hand travels in one second:
B = 2*pi / (60*60) (2)

The hands overlap in ‘x’ seconds k times, thus:
B*x-2*k*pi = S*x (3)
Substituting (1) and (2) into (3) and solving for x, we get:
x = 60*60*12*k/11 seconds (4)

Substituting k=0,1,2,…, until x
x is less than 2*12*60*60 yields k=21, so 22 crossings.

So there are 22 crossings. Note that there is NO overlap in the expected area of 11:55 because it happens at 12:00. You can spin the time setting knob on an analog clock and see this easily.

————

No, actually it happens at 11:59:59 to 12:00:00; the hour and the minute hand overlap at 11:59:59 and continue to be overlapped while moving to 12:00:00.
any way both overlap 22 times a day…

Logic needs to be applied rather than pure maths. Since there can be variations in clock accuracy, thickness etc etc. Traditional Maths is useful if you have the details to define your problem; the issue is that in this question you dont have all the details so its better to leave traditional maths out.

The fundamental thing to keep in mind is, the HOUR must corss its minute HAND every hour. (not imp when they cross in that hour).

Example: at 1:05 (within Error 1 min) they will cross.

24 Hrs imples = 24 crossings for an accurate clock!
23 for an inaccurate one.

Clock’s dial is broken into 60 small units.
Each unit represents one minute.

Minute hand moves at a speed of 60 units per hour.
That’s 1 unit per minute.

Hour hand moves at a speed of 5 units per hour.
That’s 1/12 of a unit per minute.

At analog clock position 12:00 a day starts.

The first time minute hand will overlap hour hand is
after a full hour goes by at unit position X.

From middle school math we know that:
D = S * T, where
D is distance traveled, S is speed, T is time.

To find the exact position X we realize that the
overlap condition means equality of variable T.
Starting at their initial positions, both hands
spend equal amount of time traveling to point X.
One is a bit faster. One is a bit slower.

Initial positions.
After the first hour minute hand is at 12 o’clock
position. Hour hand is at 1 o’clock position.
The distance between them is 5 units.
Where will they meet?

To answer this we say that hour hand will travel X
units past 1. And minute hand will travel 5 + X
units past 12. To meet at point X they both have to
spend the same amount of time traveling.

Solving the above equation for T we get:
T = D / S

The time it takes minute hand to get to X is:
( 5 + X ) / 1,
since its speed is 1 unit per minute.

The time it takes hour hand to get to X is:
X / 1/12,
since its speed is 1/12 unit per minute.

Now we equate the two to get proportion:
5 + X X
——- = —
1 1/12

Solving it for X gives:
5 + X = 12 * X

And that means that:
5 = 11 * X

And that means that:
X = 5 / 11 of one unit.

And that means that they first overlap after:
1 hour 5 minutes 27.27 seconds.

It is easy to see that after the second hour minute
hand will have to travel 10 + X units past 12. And
hour hand will still travel X units past 2. Giving
the second time they overlap:
2 hours 10 minutes 54.55 seconds.

Abstracting this for Nth hour we get that the
overlap point moves ( 5 * N ) / 11 past hour N.

For 11th hour overlap point is at:
( 5 * 11 ) / 11

which means it will move 5 minutes past 11. And
that is when the clock strikes 12 noon.

Conclusion.
In 12 hours from morning to noon minute hand will
overlap hour hand exactly 11 times.

Now, since during the remaining twelve hours
this behavior is repeated, we multiply 11 by two to
get 22.

The answer is 22.

There is not much cyclical in this algorithm. The
problem is easily solved if looked at from a linear
perspective and simple math.

Now, students. Extra for experts.
24-hour day.
Analog clock.
Second hand.
How many times will it overlap:
minute hand?
hour hand?

Submitted By: Saumil — October 9, 2006

-63 votes + - 24 times..

1:05, 2:10, 3:15, 4:20…..12:00, 13:05, 14:10,…..24:00.

Submitted By: Anu Thomas — October 11, 2006

+10 votes + - 22 times a day (starting from 00:00am - 23:59:60).

Submitted By: arpan — October 15, 2006

+51 votes + - The clock hands don’t overlap at 1:05, 2:10, 3:15 etc. Because, by the time the minutes hand reaches where the hours hand was, the hours hand has moved on, as it doesn’t remain fixed through the entire hour. the answer can be worked out in this way:

In every hour, as the hour hand moves across 5 minute spaces, the minute hand moves across 60. That means the minute hand gains 55 min sp. over hour hand in 60 min. Thus, between the nth and (n+1)st hr, to overlap the minute hand must gain the 5n min sp. that exists between them when the clock strikes n hrs. Apply unitary method and you get the time in which it does so i.e. 60n/11 min. So, the clock hands overlap at n hrs 60n/11 min. Obtain the values by putting m = 0 to 11, and you will find that from 00:00:00 to 23:59:59 (please note that 23:59:60 is equivalent to 24:00:00 or 00:00:00), the hands overlap 22 times.

Submitted By: Jitu — October 22, 2006

-9 votes + - Considering only minute and hour hand- during the course of each hour, minut hand overlaps the hour hand once and only once (except during the hour hand moves from 12:00 to 1:00). This maked only 24-1=23 overlaps in 24hours.

Submitted By: Debadyuti Banerjee — October 30, 2006

+12 votes + - In any circular motion, if the ratio between 2 fellow is n:1 then for each lap of the slower fellow, faster fellow overlaps him n times, except for the last lap of the slower fellow, where the number of crossings are n-1. (The race just finishes before the n th cross.

slower fellow : hour hand (short)

faster fellow : minuite hand (long)

n = 12

Total laps for hour hand = 2

total overlaps = 12+11 = 23

Submitted By: Ryan — October 30, 2006

-55 votes + - There are 24 hours each day, and with in each hour, the two hands overlap once and only once. Just 24, It’s that simple!!!

Submitted By: rdu — November 17, 2006

-27 votes + - Maybe you consider that clocks have 3 hands (hour, minute, second) then you will only get 2 overlaps 12:00 and 24:00

Submitted By: Guish — December 13, 2006

-4 votes + - The amount of times that the second hand overlaps each of the hour and the minute hand is twice a minute, i.e. two overlappings a minute (or 120 times an hour, or 2880 times a day). However, there are times where the minute hand overlaps the hour hand, and therefore the second hand overlaps them at the same time, taking away one overlapping every time this happens. We subtract one from the second hand overlappings every time the hour and minute hands overlap. Therefore, they cancel eachother out, and we are left with 2880 overlappings every day.

Submitted By: commonsense — December 23, 2006

-31 votes + - i donno how u can come to 23 or 22 hrs conclusion. all the math is not reqd. from 12 a.m to 12 p.m. how many times the hnds overlap. 12 (initial point), 1:06 ish, 2:14 ish..11:58 ish = 12. in a day it is 2 cycles i.e 24 duh!

Submitted By: bryann — February 6, 2007

-7 votes + - Let say the speed of hour hand is 1x per hr, then the speed of minute hand is 12x per hr. That means the minute hand runs faster than hour hand 11x (12x - 1x) per hour. Start from 0:00AM (first over lap), time it needs for the minute hand to over lap again is: 12x/11x or 12/11 hr. So in 24 hr the number of overlap will be 1 + 24/12/11 = 23 times.

Submitted By: Mantat — May 14, 2007

-13 votes + - 23 times (if ignore the last time12:00). I did that using the watch and loop though 24 hours. So, no way that could be wrong

Submitted By: Jennifer — July 16, 2007

-11 votes + - Here is what i think:

Assuming that hour hand, minute hand and seconds hand are of the same thickness. On a 24 hr cycle:

second hands actually move:

60 * 60 * 24 = 86400

since every 60 moves of the second hand, the minute hand only get to move once

Minute hand will have:

60* 24 = 1440;

Now the tricky part is the hour hand. It is not right to assume that hour clock only moves 24 times. Since in between each hour(example 1:00 to 2:00) hour hand actually moves the question is WHEN HE MOVES and HOW MANY movement he does before reaching the next hour. We can used this analogy below. For example 1:00 to 2:00:

distance, using the minutes hand from 1 to 2 is actually 5 minutes( or 5 ticks), using this we can also say that the hour hand will take 5 movements going to 1 to 2. In 1 hour the hour hand moves 5 times.

In 24 cycle hour hand will have:

5 * 24 = 120;

Secs = 86400

minutes=1440

hour = 120

so they will all meet 120 times in a 24 hr cycle

Submitted By: Richyinsea — August 4, 2007

-3 votes + - OK so I sat here for 24 hours and counted them.

Seriously I like the analysis of answer 5. For each hour except for the first and last in each 12 hours the hour and minute hands have to overlap exactly once. A day is 24 hours starting at midnight and ending at midnight. At those two times the hour and minute (and second hand if there is one) overlap. So those two overlaps cover the hours from midnight to 1 and from 11 to midnight when it does not otherwise overlap. However between 11 and noon and noon to 1 we have just one overlap at exactly noon since it is in the middle of the day you can’t count it twice as if the morning and afternoon were distinct unrelated entities when we are counting the overlaps in “a day.” We have 3 overlaps so far. For the other 20 hours, from 1 to 2, etc. till 10 to 11 the hour and minute hand will overlap exactly once. (The minute hand goes from 12 to 12 while the hour hand is a little past 1 or a little before 11 at the two ends and between those points for all the other hours so they have to overlap once.) So we have 23 overlaps in 24 hours counting the two midnights. If you were counting this over several consecutive days you could not count the midnight between two days twice, once for each day so it would average to 22 times. Problem says “a day” but if you interpret that to mean “each day.”

If there is a second hand then the result might be similar and depends mostly on the thickness of the hands. Every minute the second hand overlaps the minute hand once or twice or more if the minute hand is thick enough and the second hand moves in discreet jerky increments or if the hour minute hand overlap lasted longer than a minute. But if the hour and minute hands were too thin there is a chance that they would not overlap at the moment the second hand passed over the minute hand for some minutes and some of the hourly hour-minute hand overlaps would not get counted. Remember the second hand will overlap the minute hand every minute or so but we only care about when it overlaps the minute hand at the same time it is overlapping the hour hand and that only happens at 12 and every hour for the other 20 hours so assuming hour-minute overlap holds for exactly one minute the result is the same 23 overlaps.

Submitted By: Richyinsea — August 4, 2007

-3 votes + - I guess I should add as regards second hands the little clock that Google or Vista gives me has a thin second hand that moves a jerky second at a time and the hour and minute hands move smoothly and are a second (or minute) wide. The second hand can be considered to overlap the hour or minute hand once when it lands in the center of the hand or twice (in one rotation) at the edges of the hand. It will always overlap the minute hand at least once as it passes over.

Submitted By: Ayush — September 7, 2007

-2 votes + - If we consider the thickness of the hands to be very thin, and we are having 3 hands, i.e. Hour, Minute and Second. Now:

The Hour hand and Minute hand would be meeting exactly 11 times in 12 hours (Hour hand would have taken 1 clockwise round and Minute hand would have taken 12 clockwise rounds, so 12 - 1 = 11 rounds).

result: First time hour and minute hands overlap will be 12 Hours / 11 = 01:05:27.27. So at this time only hour and minute hands would be overlapping and second hand will not be any near to them. Similarly for 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th and 10th overlap of hour and minute hand the Second hand wont be any nearby. So all 3 hands (hour, minute and Second) overlap only 2 times i.e. (0:0:0 and 12:0:0).

Also we all know when we get our watches repaired, normally the repairman overlaps all the three hands to 12.

If we are considering that the second hand is not present, then the rest two overlaps 22 times in 24 hours.

There again is a catch, if we check the angles by which the hour hand and minute hand moves.

The second hand moves 6 degree in a second. In that time the minute hand will move 6/60 degrees. and the hour hand will move 6/(60*12) degrees. now taking these things in the considerations. if we check the positions of the hour and minute hand in terms of angle from the marker 12, for our first rendezvous time, i.e. 01:05:27.27 sec.

first thing that comes to my mind is that, there is fraction in the seconds. So that time can’t be measured. there will be no exact overlap. now lets calculate the angles:

1 hour 5 mins and 27 seconds = 3600 + 5*60 + 27 = 3927 seconds.

angle of hour hand = 3927 * 6/(60*12) = 32.725 degree.

angle of minute hand = 3927 * 6/60 = 392.7 degree

subtracting 360 degree from it we get - 32.7 degree.

So at 01:05:27 both hands don’t overlap. Now for 01:05:28 :

Angles : hour hand - 32.73333

minute hand - 32.8

so obviously they dont meet at 01:05:28 either.

So they overlap at 12:00 and 24:00 only. So the answer is 2 only.

Submitted By: No-it-all — September 16, 2007

-6 votes + - Assuming itâ€™s an analog, gear-driven clock, 23. The 24th overlap is technically the next day.

Submitted By: Marco Almondine — December 11, 2007

+35 votes + - The obvious guess is to write down a list of the places the hands will overlap and count them up: 1:05, 2:10, 3:15, 4:20, 5:25, 6:30, 7:35, 8:40, 9:45, 10:50, 11:55, twice around gives 24 crossings.

Unfortunately this is the wrong answer, as the hour hand must move a bit more than one hour to catch up to the minute hand. Specifically it moves 12/11 of an hour or 65.454545… minutes. Thus the correct times are:

0:00:00 (midnight)

1:05:27

2:10:55

3:16:22

4:21:49

5:27:16

6:32:44

7:38:11

8:43:38

9:49:05

10:54:33

12:00:00 (noon)

13:05:27

14:10:55

15:16:22

16:21:49

17:27:16

18:32:44

19:38:11

20:43:38

21:49:05

22:54:33

24:00:00 (which doesn’t count because it’s in the next day)

So there are 22 crossings. Note that there is NO overlap in the expected area of 11:55 because it happens at 12:00. You can spin the time setting knob on an analog clock and see this easily.

Submitted By: Andras Cser — January 23, 2008

+3 votes + - Let “S” denote the arc which the small hand travels in one second:

S = 2*pi / (60*60*12) (1)

Let “B” denote the arc which the big hand travels in one second:

B = 2*pi / (60*60) (2)

The hands overlap in ‘x’ seconds k times, thus:

B*x-2*k*pi = S*x (3)

Substituting (1) and (2) into (3) and solving for x, we get:

x = 60*60*12*k/11 seconds (4)

Substituting k=0,1,2,…, until x

x is less than 2*12*60*60 yields k=21, so 22 crossings.

Submitted By: Sikku — June 18, 2008

+0 votes + - So there are 22 crossings. Note that there is NO overlap in the expected area of 11:55 because it happens at 12:00. You can spin the time setting knob on an analog clock and see this easily.

————

No, actually it happens at 11:59:59 to 12:00:00; the hour and the minute hand overlap at 11:59:59 and continue to be overlapped while moving to 12:00:00.

any way both overlap 22 times a day…

Submitted By: Arjun Dhar — July 2, 2008

-7 votes + - Logic needs to be applied rather than pure maths. Since there can be variations in clock accuracy, thickness etc etc. Traditional Maths is useful if you have the details to define your problem; the issue is that in this question you dont have all the details so its better to leave traditional maths out.

The fundamental thing to keep in mind is, the HOUR must corss its minute HAND every hour. (not imp when they cross in that hour).

Example: at 1:05 (within Error 1 min) they will cross.

24 Hrs imples = 24 crossings for an accurate clock!

23 for an inaccurate one.

Submitted By: Casey Tompkins — November 21, 2008

+0 votes + - Correct answer: 22

Minute hand has to pass the hour hand every hour except 11:00.

Hour hand moves 1 minute every 12 minutes (60/5)

Minute hand passes the hour hand:

1 -> 1:05:25

2 -> 2:10:50

3 -> 3:16:33

4 -> 4:21:40

5 -> 5:27:12

6 -> 6:32:30

7 -> 7:37:50

8 -> 8:43:40

9 -> 9:48:30

10 -> 10:54:10

11 -> 12:00 (hands catch up at 12)

12 -> 12:00

Submitted By: UNIX Rules — December 11, 2008

+1 votes + - Clock’s dial is broken into 60 small units.

Each unit represents one minute.

Minute hand moves at a speed of 60 units per hour.

That’s 1 unit per minute.

Hour hand moves at a speed of 5 units per hour.

That’s 1/12 of a unit per minute.

At analog clock position 12:00 a day starts.

The first time minute hand will overlap hour hand is

after a full hour goes by at unit position X.

From middle school math we know that:

D = S * T, where

D is distance traveled, S is speed, T is time.

To find the exact position X we realize that the

overlap condition means equality of variable T.

Starting at their initial positions, both hands

spend equal amount of time traveling to point X.

One is a bit faster. One is a bit slower.

Initial positions.

After the first hour minute hand is at 12 o’clock

position. Hour hand is at 1 o’clock position.

The distance between them is 5 units.

Where will they meet?

To answer this we say that hour hand will travel X

units past 1. And minute hand will travel 5 + X

units past 12. To meet at point X they both have to

spend the same amount of time traveling.

Solving the above equation for T we get:

T = D / S

The time it takes minute hand to get to X is:

( 5 + X ) / 1,

since its speed is 1 unit per minute.

The time it takes hour hand to get to X is:

X / 1/12,

since its speed is 1/12 unit per minute.

Now we equate the two to get proportion:

5 + X X

——- = —

1 1/12

Solving it for X gives:

5 + X = 12 * X

And that means that:

5 = 11 * X

And that means that:

X = 5 / 11 of one unit.

And that means that they first overlap after:

1 hour 5 minutes 27.27 seconds.

It is easy to see that after the second hour minute

hand will have to travel 10 + X units past 12. And

hour hand will still travel X units past 2. Giving

the second time they overlap:

2 hours 10 minutes 54.55 seconds.

Abstracting this for Nth hour we get that the

overlap point moves ( 5 * N ) / 11 past hour N.

For 11th hour overlap point is at:

( 5 * 11 ) / 11

which means it will move 5 minutes past 11. And

that is when the clock strikes 12 noon.

Conclusion.

In 12 hours from morning to noon minute hand will

overlap hour hand exactly 11 times.

Now, since during the remaining twelve hours

this behavior is repeated, we multiply 11 by two to

get 22.

The answer is 22.

There is not much cyclical in this algorithm. The

problem is easily solved if looked at from a linear

perspective and simple math.

Now, students. Extra for experts.

24-hour day.

Analog clock.

Second hand.

How many times will it overlap:

minute hand?

hour hand?

## Leave an Answer/Comment