SOLUTION: At exactly 12 o'clock noon the hour hand of a clock begins to move at four times its normal speed, and the minute hand begins to move backward at two-thirds its normal speed. When

Question 1208690: At exactly 12 o'clock noon the hour hand of a clock begins to move at four times its normal speed, and the minute hand begins to move backward at two-thirds its normal speed. When the two hands next coincide, what will be the correct time?

Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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At exactly 12 o'clock noon the hour hand of a clock begins to move at four times its normal speed,
and the minute hand begins to move backward at two-thirds its normal speed.
When the two hands next coincide, what will be the correct time?
~~~~~~~~~~~~~~~~~~~~~~~~

The normal angular speed of the minute hand is   = 6 degrees per minute
    (one full rotation in 60 minutes).


The normal angular speed of the hour hand is   = 0.5 degrees per minute
    (one full rotation in 12 hours).


For this "mad watch" , the angular degree of the minute hand is  = -4 degrees per minute;

                       the angular degree of the hour hand is 4*0.5 = 2 degrees per minute.



The condition that the hands coincides in t minutes for this "mad watch" is

    4t + 2t = 360 degrees  (the hands rotate in opposite directions).


From this equation,

      6t = 360  --->  t = 360/6 = 60 minutes.


ANSWER.  When the two hands next coincide, the correct time will be 1:00 pm.

Solved.

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