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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]

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07 Sep 2003, 06:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Manager
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07 Sep 2003, 07:03
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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Intern
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09 Sep 2003, 10:05
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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Senior Manager
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10 Sep 2003, 12:50
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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SVP
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19 Jan 2004, 09:41
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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Senior Manager
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19 Jan 2004, 13:35
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.
_________________

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

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Senior Manager
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19 Jan 2004, 21:56
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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SVP
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19 Jan 2004, 21:59
i agree that the answer is 6 hrs.
I am just supporting wonder_gmat's theory

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19 Jan 2004, 21:59
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# A certain clock-like device has three hands and a round

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