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Clocks and their relative speeds

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data
    A watch gains 3 minutes every hour. If it is set right at 11 a.m. on February 21st, 2012, when will the hour hand of this defective watch and a correct watch be at the same position?
    A)
    9 p.m. on February 21st, 2012
    B)
    11 p.m. on February 28th, 2012
    C)
    11 a.m. on March 2nd, 2012
    D)
    11 a.m. on March 12th, 2012

    2. Relevant equations


    3. The attempt at a solution

    If the defective and the correct watch show the same time, the defective watch will have gained 12 hours, i.e., 720 minutes. Since it gains 3 minutes every hour, it will need 720/3 = 240 hours to gain 12 hours. In other words, both watches will show the same time after 10 days. Counting from 11 a.m. on February 21St, 2012, both watches will show the same time at 11 a.m. on March 2nd, 2012.

    However, the watches would show same time when the difference between them is 480 hours, so C and D should be both correct..(however the book states option C as the only correct answer)
     
  2. jcsd
  3. Nov 15, 2014 #2

    jedishrfu

    Staff: Mentor

    Did you have to factor in that 2012 is a leap year and so February has an extra day?
     
  4. Nov 15, 2014 #3
    Oops, silly me.
     
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