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Consider a well functioning clock where the hour, minute and the seconds needles are exactly at zero. How much time later will the minutes needle be exactly one minute ahead ($1/60$ th of the circumference) of the hours needle and the seconds needle again exactly at zero?

Hint: When the desired event happens both the hour needle and the minute needle have moved an integer multiple of $1/60$ th of the circumference.

  1. $144$ minutes
  2. $66$ minutes
  3. $96$ minutes
  4. $72$ minutes
  5. $132$ minutes 
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Please correct me if I am wrong. I think answer is (b) 66 min.

 

Reason :

Here it is given current time is 12:00 pm. After 66 min , time will be 1:06 p.m . So , hour's hand will be at 1 and minute's hand will be exactly 1 place ahead.
Answer:

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