Subject: geometry 10th grade
Name: kate
Who is asking: Student
Level: Secondary

While repairing a watch, a jeweler removed the hands and inadvertently replaced the hour hand on the min. spindle and vice versa. he set the hands to read 2:00pm, which was the correct local time when I picked up the watch. A few minutes later, I noticed that the hands were taking goofy positions. What was the first time thereafter that the watch would show the correct local time?

I drew pictures of clock with the hour hand on two and minute hand at zero like a regular clock. instead of having the minute hand go around for each hour, the hour hand did. this process resulted in the times:

when its 2pm correct time, the wrong clock is 2pm
correct time: 3pm, wrong clock is: 2:05
correct time: 4pm, wrong is: 2:10
correct time: 5pm, wrong is: 2:15

I think you've caught along with the pattern by now, using this pattern I got that the first time was somewhere around 24 hours later that the wrong time was equal to the the right time. I showed this solution to my teacher but he said the problem wasn't anything that easy, the answer is somewhere from 65 to 66 minutes after two oclock. But, I don't know how to get to that answer, thats why I need help - to find strategy! This problems due this monday, so help quick!thanks

Hi Kate,

I don't really understand. Watches don't work as described: when the minute hand points to 10 after, the hour hand is 1/6 of the way between the 12 and the 1. Doesn't a meaningful time occurs only when the hands cross?

The problem reminds me of another one, its solution may help.

PROBLEM: The hour and minute hands are together at exactly twelve noon, and they come together again sometime between 1:00 and 2:00. At what instant does this next crossover occur?

ANSWER: To understand the problem you need to realize that the hands cross at regular intervals, say every x minutes. (Just ignore the numbers and you will see that there is nothing special about 12:00 - no matter when the hands are together, they will be together again x minutes later.) In each twelve-hour period there are exactly 11 cross-overs; thus x = 12*60/11 = 65.4545..., which is about 1h 5m 27s; i.e. the first cross-over after 12:00 occurs 27 seconds after 1:05.



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