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October 11th, 2018, 08:42 PM   #1
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how to solve this confusing clock problem?

Q) At what time between 4-5, both hands of clock right angle side.

A) The minute hand moves 1/60th of the way around every minute. So it moves 6° per minute.

The hour hand moves 1/12th = 30° every hour. So it moves 0.5° every minute.

At 4 o'clock, taking the 12-position to be 0°, the minute hand will be at 0° and the hour hand will be at the 4-position = 120°.

After n minutes, the minute hand will be at (0+6n)° = (6n)° and the hour hand will be at (120+0.5n)°.

We want the difference to be 90°, so;

120 + 0.5n - 6n = 90

30 = 5.5n

n = 5.454545

So the time will be 4:05:27

My friend got two answers: 4h60/11 min and 4h420/11 min.

I am confused.
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October 12th, 2018, 12:46 AM   #2
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30/5.5 = 60/11

Your friend also found the time when the minute hand is 90° beyond the hour hand.
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October 12th, 2018, 03:33 AM   #3
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Quote:
Originally Posted by skipjack View Post

Your friend also found the time when the minute hand is 90° beyond the hour hand.
What is the meaning of "beyond the hour hand"?
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October 12th, 2018, 08:20 AM   #4
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As the minute hand is moving much faster than the hour hand, it will overtake it and reach an angle of 90 degrees with it before 5 o’clock.
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