# Clock problem

• Sep 7th 2012, 06:08 AM
rcs
Clock problem
At what time between 2 and 3 o'clock are the hands of the clock opposite each other?

please help me on this... kind of dont understand the phrase "the hands of the clock opposite each other" .
i need your guide on this.

thanks
• Sep 7th 2012, 06:44 AM
Prove It
Re: Clock problem
Quote:

Originally Posted by rcs
At what time between 2 and 3 o'clock are the hands of the clock opposite each other?

please help me on this... kind of dont understand the phrase "the hands of the clock opposite each other" .
i need your guide on this.

thanks

It means, at what time in between 2:00 and 3:00 are the hands pointing directly opposite each other (making a straight angle)?
• Sep 7th 2012, 07:04 AM
rcs
Re: Clock problem
Quote:

Originally Posted by Prove It
It means, at what time in between 2:00 and 3:00 are the hands pointing directly opposite each other (making a straight angle)?

According to the book where i got this problem, the answer is 43 and 7/11 min. after 2 oclock. I dont know how they got it because it was not manipulated or shown
• Sep 7th 2012, 07:17 AM
Prove It
Re: Clock problem
Here's how I would try to do it. In one hour, the hour hand will move from the 2 to the 3, and the minute hand will do a complete rotation from the 12 back to the 12. It should be obvious that the minute hand needs to at least get to the 8 (where it will be opposite the 2) and can not get to the 9 (opposite the 3).

To get to the 8 takes 40 minutes. 40 minutes is 2/3 of an hour, so the hour hand will be 2/3 of the way between 2 and 3.

Now think "every minute, the hour hand will move 1/60 of the way between 2 and 3, and the minute hand will move 1/5 of the way between 8 and 9".

See if you can continue.
• Sep 7th 2012, 07:45 AM
emakarov
Re: Clock problem
Attachment 24725

I suggest denoting by x the fraction of the hour that passed after 2 o'clock. Thus, $\displaystyle \frac{8}{12}<x<\frac{9}{12}$.
• Sep 8th 2012, 05:15 AM
rcs
Re: Clock problem
thanks Sir ProveIt ....