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        riddles >> easy >> Analog Clock II
        
(Message started by: osirusblue on Aug 29^th, 2007, 12:05am)

Title: Analog Clock II
Post by osirusblue on Aug 29^th, 2007, 12:05am

Is my thinking flawed that there are 11 times the hands of clock will be on top of each other?

I had been told that it was 10 but for the life of me I can only make it make sense if I say 11.

1:05
2:10
3:15
4:20
5:25
6:30
7:35
8:40
9:45
10:50
11:55

No? I understand the AM/PM changes it a bit but if we say just within a 12 hour period?

Sorry if I should have hidden any of this... first time caller here.

Title: Re: Analog Clock II
Post by mikedagr8 on Aug 29^th, 2007, 12:08am

No it's 11.

Title: Re: Analog Clock II
Post by towr on Aug 29^th, 2007, 12:29am

The little hand makes one round, the large hand makes 12, so it has to pass the little one 11 times.

Title: Re: Analog Clock II
Post by osirusblue on Aug 30^th, 2007, 12:11am

That's what I thought. Thank you both. Now, I must be beating somebody for making me think I was losing my mind!

Cheers!

Title: Re: Analog Clock II
Post by rmsgrey on Aug 30^th, 2007, 2:26am

on 08/29/07 at 00:05:44, osirusblue wrote:

1:05
2:10
3:15
4:20
5:25
6:30
7:35
8:40
9:45
10:50
11:55

Those times aren't exact - it's actually 1:05:27.2727... between overlaps - in particular, the last time on the list should actually be 12:00 again.

Depending on the exact phrasing of the question, considering the 12 hour interval from noon to midnight (or midnight to noon), the number of times the two hands coincide could be 10, 11, or 12, according to whether you count one, both, or neither 12 o'clocks.