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A certain clock-like device has three hands and a round

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A certain clock-like device has three hands and a round [#permalink] New post 07 Sep 2003, 05:37
A certain clock-like device has three hands and a round face. The first hand rotates at the speed of 1 rotation per hour, the second—1/2 rph, and the third—1/3 rph. Initially, all the hands are set on "a 12 o'clock position" and start to rotate. In how many minutes will all the hands meet again, the first time after the start? In which position?
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 [#permalink] New post 07 Sep 2003, 06:03
In 360 minutes???? Given 6 is the first integer evenly divisible by 2 AND 3. So at the 6th hour they'll all meet at spot where they started.

If they don't have to meet at THE initial spot then they'll all meet at approximately 110 minutes.

I'm still iffy about my answer. Let's see what others have to say...
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 [#permalink] New post 09 Sep 2003, 09:05
wonder_gmat wrote:
In 360 minutes???? Given 6 is the first integer evenly divisible by 2 AND 3. So at the 6th hour they'll all meet at spot where they started.

If they don't have to meet at THE initial spot then they'll all meet at approximately 110 minutes.

I'm still iffy about my answer. Let's see what others have to say...


How did you get 110 min?
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 [#permalink] New post 10 Sep 2003, 11:50
3 Hands: A,B and C
A:1 rotation/h
B:1/2 rotations/h
C:1/3 rotations/h

After they start together, in each hour hand B rotates 1/6 more than C does. So B passes
over C every 6 hours.
In each hour hand A rotates 1/2 more than hand B does. Likewise, hand A passes
over B every 2 hours.

The first time in which A passes over B at the same moment
B passes over C is at 6 hours.

And the position is the same in which they started.



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 [#permalink] New post 19 Jan 2004, 08:41
F completes one rotation in 1 hr
S completes 1/2 in 1 hr
T completes 1/3 in 1 hr
together they complete
1+1/2+1/3 rotations = 11/6 rotation in one hr
So they meet after 60 * 11/6 minutes = 110
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 [#permalink] New post 19 Jan 2004, 12:35
6 hours...

Opted for the traditional,time consuming method..

After 1 hour

1 ----- 12
2 ------ 6
3 ------ 4

After 2 hours

1 ----- 12
2 ------ 12
3 ------ 8

After 3 hours

1 ----- 12
2 ------ 6
3 ------ 12

After 4 hours
1 ----- 12
2 ------ 12
3 ------ 4

After 5 hours
1 ----- 12
2 ------ 6
3 ------ 8

After 6 hours
1 ----- 12
2 ------ 12
3 ------ 12


Vivek.
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"Start By Doing What Is Necessary ,Then What Is Possible & Suddenly You Will Realise That You Are Doing The Impossible"

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 [#permalink] New post 19 Jan 2004, 20:56
anandnk wrote:
F completes one rotation in 1 hr
S completes 1/2 in 1 hr
T completes 1/3 in 1 hr
together they complete
1+1/2+1/3 rotations = 11/6 rotation in one hr
So they meet after 60 * 11/6 minutes = 110


I do not think that this is correct logic. 11/6 is just the total of the rotations finished by all the three hands IN ONE HOUR. But all three will not be in the same position. In ONE HOUR, the first hand will again be in 12 oclock position, the secon will be on 6 and the third ahnd would be on 4. They will not conside in any way.
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 [#permalink] New post 19 Jan 2004, 20:59
i agree that the answer is 6 hrs.
I am just supporting wonder_gmat's theory
  [#permalink] 19 Jan 2004, 20:59
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A certain clock-like device has three hands and a round

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