Question 1204412: At exactly 12 o'clock noon the hour hand of a clock begins to move at twice its normal speed, and the minute hand begins to move backward at half its normal speed. When the two hands next coincide, what will be the correct time?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850)   (Show Source):
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Answer by ikleyn(52903)  (Show Source):
You can put this solution on YOUR website! .
At exactly 12 o'clock noon the hour hand of a clock begins to move at twice its normal speed,
and the minute hand begins to move backward at half its normal speed.
When the two hands next coincide, what will be the correct time?
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The normal angular speed of the hour hand is 360/12 = 30 degrees per hour;
so, its doubled speed is 60 degree per hour.
The normal angular speed of the minute hand is 360 degrees per hour;
so, half of its speed is 180 degrees per hour.
Thus, in this problem, the hour hand rotates anti-clockwise at 60 degrees per hour;
the minute hand rotates clockwise at 180 degrees per hour.
They approach each other at the rate of (60+180) = 240 degrees per hour.
So, they coincide next time in 360/240 = 1 = 1.5 hours = 90 minutes.
When the clock hands coincide next time, the correct time is 1:30 pm. ANSWER
Solved.
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