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 of the day that this occurs?

Let X be the number of the hour Let t be the minutes passed the hour So the time is X:t

Let h be the position of the hour hand Let m be the position of the hour hand

h = X + t/60 m = t/5

Now time find where the two hands cross, we need to know when h = m t/5 = X + t/60 12t = 60x +t 11t = 60x t = 60X/11 For the first crossing, X =1 t = 60/11 t =5.45 minutes t = 5 minutes and 27 seconds (note 27/60 is about 0.45) So the first crossing is at 1:05:27

The second crossing is X =2. t = 120/11 t= 10 minutes and 55 seconds Second crossing at 2:10:55

By setting X to each hour you can solve for the remaining 9 crossing.

I got the same answer, I think the same way. The hands cross 11 times in 12 hours = every 12/11 of an hour = every 1:05:27.27... [ March 12, 2006: Message edited by: Stan James ]

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