phillip1882 Posted April 24, 2012 Share Posted April 24, 2012 three men have watches that are all oddly set. the first man's watch gets five minutes faster every hour. he reports the current time as 3:15 pm. the second man's watch counts backward one third as fast. he reports the time as 2:30 pm. the third man's watch moves at a standard pace, but marks 20 hours in a day, and 100 minutes in an hour. he reports the time as 5:45 pm. will it ever be the case that the three men report the same time? Link to comment Share on other sites More sharing options...
questionposter Posted April 25, 2012 Share Posted April 25, 2012 (edited) three men have watches that are all oddly set. the first man's watch gets five minutes faster every hour. he reports the current time as 3:15 pm. the second man's watch counts backward one third as fast. he reports the time as 2:30 pm. the third man's watch moves at a standard pace, but marks 20 hours in a day, and 100 minutes in an hour. he reports the time as 5:45 pm. will it ever be the case that the three men report the same time? There's probably some way to set it up a system of equations. time=3:15+(1+5/60)x time=2:30-(1+1/3)x (5/6)time=(1+6/10)x And I'll get back to it later. 3:15+(1+5/60)x = 2:30-(1+1/3)x [(1+6/10)x]/(5/6) = 2:30-(1+1/3)x Edited April 25, 2012 by questionposter Link to comment Share on other sites More sharing options...
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