Results 1 to 9 of 9
Like Tree6Thanks
  • 1 Post By romsek
  • 1 Post By romsek
  • 1 Post By Soroban
  • 1 Post By SlipEternal
  • 1 Post By Soroban
  • 1 Post By SlipEternal

Math Help - Watch gaining and losing time

  1. #1
    Senior Member
    Joined
    Apr 2006
    Posts
    290

    Watch gaining and losing time

    My watch (which is a 12 hour watch) GAINS 3 mins every 2 hours.

    a) I set my watch to the correct time at noon on 1st January. If I don't reset it, when will it next show the correct time?

    I got:

    real time vs gain time
    2hrs vs 3 mins gain
    40hrs vs 60 mins (x20)
    960 hrs vs 24 hrs (x24)
    = 40 days (divided 960 hrs by 24hrs)

    Is this correct please for a?

    b) Mrs Warma's watch (also 1 12 hour watch LOSES 5 mins every 2 hours. She also sets her watch to the correct time at noon on 1st January. When will our two watches next show the same time?

    I got:

    real time vs lose time
    2hrs vs 5 mins lose

    I am stuck after this....


    c) When will our watches next show the same, correct time?

    I am stuck with this

    Thanks for help in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    1,533
    Thanks
    557

    Re: Watch gaining and losing time

    Quote Originally Posted by Natasha1 View Post
    My watch (which is a 12 hour watch) GAINS 3 mins every 2 hours.

    a) I set my watch to the correct time at noon on 1st January. If I don't reset it, when will it next show the correct time?

    I got:

    real time vs gain time
    2hrs vs 3 mins gain
    40hrs vs 60 mins (x20)
    960 hrs vs 24 hrs (x24)
    = 40 days (divided 960 hrs by 24hrs)

    Is this correct please for a?

    b) Mrs Warma's watch (also 1 12 hour watch LOSES 5 mins every 2 hours. She also sets her watch to the correct time at noon on 1st January. When will our two watches next show the same time?

    I got:

    real time vs lose time
    2hrs vs 5 mins lose

    I am stuck after this....


    c) When will our watches next show the same, correct time?

    I am stuck with this

    Thanks for help in advance.
    Not quite. It's a 12 hr watch so it will be correct again in 12 hrs, not 24 as you've used. So it will be 20 days not 40.

    b)

    \mbox{in minutes your watch shows }m1=t(1+3/120)=t(1+1/40)=\frac{41}{40}t\mbox{ minutes.}

    \mbox{her's shows }m2=t(1-5/120)=t(1-1/24)=\frac{23}{24}t\mbox{ minutes.}

    \mbox{The watches are able to display } t \bmod 720  \mbox{ minutes uniquely.}

    \mbox{So we want to solve for }\left(m1 \bmod 720\right)=\left(m2 \bmod 720\right)

    see if you can work it from there.

    c) you need to calculate how often the lady's watch is correct similar to how you did in (a). Then find the least common multiple of the three times, how often your watch is right, how often hers is, how often they are the same.
    Last edited by romsek; December 23rd 2013 at 03:26 AM.
    Thanks from Natasha1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2006
    Posts
    290

    Re: Watch gaining and losing time

    i don't understand what mod is, never done this at school
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    1,533
    Thanks
    557

    Re: Watch gaining and losing time

    a mod b = remainder of a/b

    for example (25 mod 7) = 4 because 25 - 3x7 = 4

    I'll give you a hint solving this.

    Try solving \frac{41}{40}t-720=\frac{23}{24}t
    Thanks from Natasha1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539

    Re: Watch gaining and losing time

    Hello, Natasha1!

    My watch (which is a 12-hour watch) GAINS 3 mins every 2 hours.

    a) I set my watch to the correct time at noon on 1st January.
    If I don't reset it, when will it next show the correct time?

    Your watch gains 1.5 minutes every hour.

    To show the correct time again, it must gain 12 hours = 720 minutes.

    . . \frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{1.5\text{ min.}} \:=\:480\text{ hours} \:=\:20\text{ days}

    It will show the correct time at noon on 21 January.

    Thanks from Natasha1
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,391
    Thanks
    516

    Re: Watch gaining and losing time

    Quote Originally Posted by Natasha1 View Post
    b) Mrs Warma's watch (also 1 12 hour watch LOSES 5 mins every 2 hours. She also sets her watch to the correct time at noon on 1st January. When will our two watches next show the same time?
    Your two watches will show the same time at the moment when your watch gains t minutes and Mrs. Warma's watch loses 12 hours minus t minutes. So, let h be the number of periods of two hours that must pass before your watch gains t minutes (and Mrs. Warma's watch loses 720-t minutes). Then 3h = t and 5h = 720-t. This means t=3h and t=720-5h. So, 3h = 720-5h shows that 8h = 720, and h = 90. That means their watches will show the same time every 2h = 180 hours. So, their watches will show the same time midnight, January 9th (that is the midnight that falls between the 8th and the 9th).
    Thanks from Natasha1
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539

    Re: Watch gaining and losing time

    Hello again, Natasha1!

    b) Mrs Warma's watch (a 12-hour watch) LOSES 5 mins every 2 hours.
    She sets her watch to the correct time at noon on 1st January.
    When will our two watches next show the same time?

    Your watch gains 1.5 minutes every hour.
    To show the correct time again, your watch must gain 12 hours = 720 minutes.
    . . \frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{1.5\text{ min.}} \:=\:480\text{ hours} \:=\:20\text{ days}

    Her watch loses 2.5 minutes every hour.
    To show the correct time again, her watch must lose 12 hours = 720 minutes.
    . . \frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{2.5\text{ min.}} \:=\:288\text{ hours} \:=\:12\text{ days}

    The two watches will agree in: LCM(20,12) = 60 days . . . Noon on March 30.
    Thanks from Natasha1
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,391
    Thanks
    516

    Re: Watch gaining and losing time

    Quote Originally Posted by Soroban View Post
    Hello again, Natasha1!


    Your watch gains 1.5 minutes every hour.
    To show the correct time again, your watch must gain 12 hours = 720 minutes.
    . . \frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{1.5\text{ min.}} \:=\:480\text{ hours} \:=\:20\text{ days}

    Her watch loses 2.5 minutes every hour.
    To show the correct time again, her watch must lose 12 hours = 720 minutes.
    . . \frac{720\text{ min.}}{1} \times \frac{1\text{ hr.}}{2.5\text{ min.}} \:=\:288\text{ hours} \:=\:12\text{ days}

    The two watches will agree in: LCM(20,12) = 60 days . . . Noon on March 30.
    This is the answer to part (c), not part (b).
    Thanks from Natasha1
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Apr 2006
    Posts
    290

    Re: Watch gaining and losing time

    Many thanks to all of you . Merry Christmas everyone!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: November 14th 2009, 01:02 PM

Search Tags


/mathhelpforum @mathhelpforum