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## Question

x by mistake dropped his two watches in the water.As a result one clock is gaining 2 seconds in each hr while other is losing 2.5 seconds an hour.Now, if their watches show the same time,how long will it be before they show the same time again.

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This sounds more like a homework problem than a logic puzzle. The solution is simple:

HINT]recognize the differential time between watches has greater significance than differential between each watch and real time.

I hope the hint gives you sufficient to see the solution method but I won't give you the answer.

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We could make this a little more interesting so I will suggest a part B:

Assume that this occurred at noon today 11/20/2011. Give date and time that the two watches once again are coordinated.

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I agree this sounds more like a homework problem but it doesn't look like they're coming back. Interesting results with my stab at it

4.5 seconds/hour would take 400 days to have the total of the watches change by 12 hours. Coincidentally I'll be wearing my santa hat next year and having a nice midday lunch when that occurs.

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I agree this sounds more like a homework problem but it doesn't look like they're coming back. Interesting results with my stab at it

4.5 seconds/hour would take 400 days to have the total of the watches change by 12 hours. Coincidentally I'll be wearing my santa hat next year and having a nice midday lunch when that occurs.

Assuming you wear your Santa's hat on 12/25/2012, Christmas day, then your date is in error!
is the fact that 2012 is a leap year, hence 366 days long.
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Assuming you wear your Santa's hat on 12/25/2012, Christmas day, then your date is in error!

is the fact that 2012 is a leap year, hence 366 days long.

Pity...it would have been an interesting coincidence.

Edited by Morningstar

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