SOLUTION: Roger started a trip to the country between 8 am and 9 am when the hands of the clock were together. He arrived at his destination between 2 pm and 3 pm, when the hands of th

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Question 1124205: Roger started a trip to the country between 8 am and 9 am
when the hands of the clock were together. He arrived at
his destination between 2 pm and 3 pm, when the hands of
the clock were exactly 180° apart. How long did he travel?

Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

The hour hand makes 1 revolution in the time that the minute hand
makes 12 revolutions, so since 12-1 = 11, the hands of a clock are 
together 11 times in 12 hours.

The hands of a clock are together at 12 o'clock.

We divide 12 hours by 11 and get 1.0909090909... hours between
successive times when the hands are together.

So the hands are together these ll times 

12.00000... o'clock.
1.090909... o'clock
2.181818... o'clock
3.272727... o'clock
4.363636... o'clock
5.454545... o'clock
6.54545.... o'clock
7.63636.... o'clock
8.72727.... o'clock
9.81818.... o'clock
10.90909... o'clock

[Then if we add 1.090909... again we're back to 12 o'clock.]

So Roger started at 8.72727... o'clock AM.

Now we need to figure out when he arrived.
The hands were together at 2.181818... o'clock and
again at 3.272727... o'clock.  Exactly halfway between
those two time, the hands were 180° apart, so we average
those two times:

    2.181818...
   +3.272727...
   ------------
    5.454545...

Then divide that by 2 and get 2.72727... o'clock PM.  But 
on a 24 hour clock we would add 12 and get 14.72727... hours.

So he started at 8.72727... hours on a 24-hour clock and
arrived at 14.72727... hours on a 24-hour clock

So he traveled for 14.72727... minus 8.72727... or 6 hours.

Edwin

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