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# Clock

21 replies to this topic

### #11 dak1530

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Posted 06 November 2007 - 09:32 PM

it is very simple and as usual people are over complicating the issue. at 1 hr and 5 mins both the hour hand and the minute hand are on the 1 because that is 1 hr and 5 minutes after 12. now for it to be exact the second had has to be on the 12 at that exact second for it to be 1 hr and 5 minutes past 12 on a complete rotation. therefore the second hand is 30 degrees exactly from both the hour hand and the minute hand which are both exactly on each other.
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### #12 OldDogCage

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Posted 08 November 2007 - 09:00 PM

its 1:05 and five seconds, 30 degrees
then 2:10 and 10 seconds, 60 degrees
next 3:15 and 15 seconds, 90 degrees.
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### #13 OldDogCage

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Posted 09 November 2007 - 02:13 AM

oops, misread, original answers sound correct now. Isn't this more of a math, than logic puzzle
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### #14 al_rawi

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Posted 21 November 2007 - 06:09 PM

I started to write this asking a question, but managed to answer it while typing.

I was doing it mathematically, and it works like this:
where
2 * pi is a 360 degree angle in radians
t is the time in milli seconds

so we want
hours angle = minutes angle +/- 2 * pi * X
(+/- 2 * pi * X is used to denote that we don't care how many more or less times the minutes turn full turns extra than the hours one, note that X can be an integer between 0 and 11 inclusive)

so

2 * pi * t / (12 * 60 * 60 * 1000) = 2 * pi * t / (60 * 60 * 1000) +/- 2 * pi * X

solving for t in terms of X yields:
t = 12 * 60 * 60 * 1000 * X / 11

you can get the time for t at X = 0, 1, 2, 3, .. , 11 to figure out the time in milliseconds from 12

of course if you don't like pis and radians, you could have used degrees or nothing (parts of a circle) to simplify as it cancels out from the equation anyway.

OK, so it is a bit more confusing but it's the way I went
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### #15 jec

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Posted 05 December 2007 - 02:18 AM

here it is since the hour and minute hand are moving calculate the time it takes for the minute hand to catch the hour hand

second hand: 6 degree/sec
minute hand: 1/10 degree/sec
hour hand: 1/120 degree/sec
for the first time t*1/10-(t*1/120+30deg)=0 solve for t
t=3600/11 = 327.2727 seconds
it will take 2*t for second time or 654.5454... etc
angle of minute hand=hour hand= t*1/10
angle of second hand= t*6
subtract the difference to get the angle between
the here is the table:
N t Min Sec Ang SH Ang MH Angle
0 0.00 0.00 0.00 0.00 0.00 0.00
1 327.27 5.45 27.27 163.64 32.73 130.91
2 654.55 10.91 54.55 327.27 65.45 261.82
3 981.82 16.36 21.82 130.91 98.18 32.73
4 1309.09 21.82 49.09 294.55 130.91 163.64
5 1636.36 27.27 16.36 98.18 163.64 65.45
6 1963.64 32.73 43.64 261.82 196.36 65.45
7 2290.91 38.18 10.91 65.45 229.09 163.64
8 2618.18 43.64 38.18 229.09 261.82 32.73
9 2945.45 49.09 5.45 32.73 294.55 261.82
10 3272.73 54.55 32.73 196.36 327.27 130.91
11 3600.00 60.00 0.00 0.00 0.00 0.00
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### #16 woon

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Posted 17 March 2008 - 10:49 AM

I calculate it like the below way:

For hour hand, 60 minutes = 30 degrees (1/12 of 1 round), so the rate = 30 / 60 = 1/2 deg per minute

For minute hand, 60 minutes = 360 deg ( 1 full round), so the rate = 360 / 60 = 6 deg per minute

let t = the time we interest,

so in any time the formula will be:

for hour hand = 6 * t - n * 360 ( n = whole number, so that if the number of round > 1, just deduct accordingly so that the angle within 0 to 360)

and for minue hand = 1/2 * t

when hour hand and minute hand overlay:

6t - n(360) = t/2

so, (11/2)t = n(360)

t = (2/11) * n(360), where n = 0, 1, 2, 3, 4, ....

t = 0/11 min, 720/11 min, 1440/11 min, 2160/11 min ...... etc

since we talking about the time when around 1 hour and 5 minuts, the answer for t should be very near to 65 minutes, which is the second one, t = 720/11 min

or t = 65 min and 27.273 sec

the angle by that time is 720/11 * 1/2 = 32.73 deg, and by that time, the second hand is making a degree of ( 6 * 27.273) = 163.64 deg,

and the angle formed is 163.64 - 32.73 = 130.9 deg.

Correct? I didn't see the answer from Spoilers!
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### #17 ALFRED

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Posted 20 March 2008 - 09:03 PM

Because the hour hand moves forward as the minute hand moves, they only cross paths 11 times in a 12 hour period.

12 hours x 60 min x 60 sec = 43,200 seconds.

43,200/11 = 3927 & 3/11 seconds = 1 hour, 5 min, 27 & 3/11 seconds.

So the second had will be at 27 & 3/11 seconds, or approx 27.2727. (By the way, this is exactly 5/11 of the way around the clock face, which will help us with the next problem).

As to the angle formed between the hour/minute hands compared to the second hand, we need only realize that the hour and minute hand have traveled exactly 1/11 of the circumference of the clock face, while the second hand is exactly 5/11 of the same circumference. Since the difference is 4/11, and a complete circle is 360 degrees, the distance is 360 x 4 / 11 = 130 & 10/11 or approx 130.90909 degrees.

VERY NICE EXPLANATION.
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### #18 bobbyb

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Posted 24 April 2008 - 07:00 AM

Because the hour hand moves forward as the minute hand moves, they only cross paths 11 times in a 12 hour period.

12 hours x 60 min x 60 sec = 43,200 seconds.

43,200/11 = 3927 & 3/11 seconds = 1 hour, 5 min, 27 & 3/11 seconds.

So the second had will be at 27 & 3/11 seconds, or approx 27.2727. (By the way, this is exactly 5/11 of the way around the clock face, which will help us with the next problem).

As to the angle formed between the hour/minute hands compared to the second hand, we need only realize that the hour and minute hand have traveled exactly 1/11 of the circumference of the clock face, while the second hand is exactly 5/11 of the same circumference. Since the difference is 4/11, and a complete circle is 360 degrees, the distance is 360 x 4 / 11 = 130 & 10/11 or approx 130.90909 degrees.

I agree with you up till the last part, if it takes an hour five minutes and approx. 27.2727 seconds, then you have to apply that to the one o'clock hour also. so the next time the hour and minute hands would overlay would be 2 hours ten minutes and approx. 54.5454 seconds from noon/midnight (0:0:0). If you are talking about the angle as the clock moves, from the ten minute mark to the ~54.5 second mark is between 253 and 254 degrees.
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### #19 beatbox

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Posted 03 May 2008 - 01:09 PM

My solution:

The hour hand moves at 1/2deg per sec and the minute hand at 6deg/sec.

The minute hand will move 360 more degrees than the hour hand when they meet.

calling minutes x we see: 1/2 x+360=6x so (11/2)x=360 and x=720/11 minutes.

The degree covered according to the hour hand is .5x or (1/2)*(720/11) = 360/11

Since 720/11=65minutes and 5/11 seconds, the second hand is 5/11(360) or 1800/11

1800/11-360/11=1440/11 degrees between the hour and second hand.
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### #20 -x-babiiee-hezzahh-x-

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Posted 11 May 2008 - 09:15 PM

if its 1 hour 5 mins after 12noon exactly its 13:05, theres no mention of the second hand...you are overcomplicationg it

&& the angle is 30degrees i think because a full circle is 360 degrees so you quarter it to make 90 and devide by 3...making the answer to the riddle 13:05 and 30 degrees.....thereis no need for long sums or pi...=S

Edited by -x-babiiee-hezzahh-x-, 11 May 2008 - 09:16 PM.

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