30. The Case of the Smithsonian Clocks Two friends, Arthur and Robert, set their clocks right at exactly 12 noon. One clock loses ten seconds an hour, and the other gains ten seconds an hour. When will the two clocks show the same time again? Added 26 March 2010 Show Solution Solution: Let t be the number of real hours elapsed. The slow clock loses 10 seconds each hour, so its displayed time in seconds is (3600–10)t=3590t. The fast clock gains 10 seconds each hour, so its displayed time is (3600+10)t=3610t. On a 12-hour dial the two readings agree whenever their difference is a multiple of 12 hours (43 200 seconds), i.e. when3610t−3590t=20t=43 200·n.For the first nonzero coincidence take n=1:t=43 200/20=2160 hours=90 days.Thus they will next show the identical time after 90 days (2160 hours).
Comments (0)
No comments yet. Be the first!
Add a Comment or Suggest an Answer