Results 1 to 10 of 10

Math Help - challenging rate problem

  1. #1
    Member
    Joined
    Apr 2010
    From
    dfsdfdf
    Posts
    160

    challenging rate problem

    Dear Sir,
    I would apppreciate very much if someone can help me in the below problem.
    Thanks.

    Tom and John were jogging along a circular track surrounding a pond. The track measured 640 m. If they started from the same place and jogged in the same direction, Tom would take 16 minutes to catch up with John. If they jogged in the opposite direction, they would meet every 4 minutes. How long did Tom take to jog one round of the track?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2011
    From
    United States
    Posts
    32

    Re: challenging rate problem

    @kingman

    My GUESS(i'm not completely sure):

    Algebraically:
    640meters; Distance or circumference of the track
    4mintues; How much time passes when Tom is passed by John
    640meters/4mintues = 120meters; Distance Tom travels when he is passed by John
    640meters/120meters = 8laps; Number of 120m segments
    8laps * 4mintues = 32minutes; How much time it takes Tom to complete the track

    NOTE: Units are incorrect.

    Graphing:
    challenging rate problem-tom-john.jpg


    @All

    What are the three ways to solve math problems? I forgot.
    Last edited by ArcherSam; September 3rd 2011 at 08:16 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,174
    Thanks
    74

    Re: challenging rate problem

    Since Tom must jog 1 lap (640 m) more than John before catching up,
    then the problem is same as Tom starting off 1 lap (640m) behind John:
    Code:
                 (640)            John........@ a mpm.............>16 min
     
    Tom.......................@ (160 - a) mpm.....................>16 min
    Since they meet every 4 min if starting in different directions, then
    combined speed is 640/4 = 160 mpm (meters per minutes):
    then, let John's speed = a mpm: then Tom's speed = 160 - a mpm.

    Using the basic speed = distance / time formula, then for Tom:
    (16a + 640) / (160 - a) = 16
    Solve to get a = 60 : so Tom's speed = 160 - 60 = 100 mpm.

    So time for Tom to jog 1 lap = 640 / 100 = 6.4 min = 6 min 24 sec.
    Last edited by Wilmer; September 3rd 2011 at 05:29 PM. Reason: Change 6:40 to 6:24
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,174
    Thanks
    74

    Re: challenging rate problem

    Quote Originally Posted by ArcherSam View Post
    What are the three ways to solve math problems? I forgot.
    LSP: Listen, Study, Practice !

    How did you get this: "640meters/4mintues = 120meters" ? 640 / 4 = 160 ; no coffee yet?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Apr 2010
    From
    dfsdfdf
    Posts
    160

    Re: challenging rate problem

    Quote Originally Posted by Wilmer View Post
    Since Tom must jog 1 lap (640 m) more than John before catching up,
    then the problem is same as Tom starting off 1 lap (640m) behind John:
    Code:
                 (640)            John........@ a mpm.............>16 min
    
    Tom.......................@ (160 - a) mpm.....................>16 min
    Since they meet every 4 min if starting in different directions, then
    combined speed is 640/4 = 160 mpm (meters per minutes):
    then, let John's speed = a mpm: then Tom's speed = 160 - a mpm.

    Using the basic speed = distance / time formula, then for Tom:
    (16a + 640) / (160 - a) = 16
    Solve to get a = 60 : so Tom's speed = 160 - 60 = 100 mpm.

    So time for Tom to jog 1 lap = 640 / 100 = 6.4 min = 6 min 40 sec.

    thanks but is'nt 6.4 mins equals 6 mins and 24 secs.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Apr 2010
    From
    dfsdfdf
    Posts
    160

    Re: challenging rate problem

    Can you explain how you reason out that Tom must jog 1 lap (640 m) more than John before catching up.
    They started at the same point ( for eg point A) and I wonder whether they meet at same point A after 16 mins.
    thanks
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,174
    Thanks
    74

    Re: challenging rate problem

    Quote Originally Posted by kingman View Post
    thanks but is'nt 6.4 mins equals 6 mins and 24 secs.
    Yes, my bad; edited.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,174
    Thanks
    74

    Re: challenging rate problem

    Quote Originally Posted by kingman View Post
    Can you explain how you reason out that Tom must jog 1 lap (640 m) more than John before catching up.
    They started at the same point ( for eg point A) and I wonder whether they meet at same point A after 16 mins.
    thanks
    A little "difficult" to "see", I agree; here's proof:
    16 min @ 100 = 1600 m - 2*640 = 320
    16 min @ 60 = 960 - 640 = 320
    So Tom catches up at point 320 m , having gone around 1 extra "640".

    Anyhow, YOU draw a circle and simulate what's happening; you'll see!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2011
    Posts
    71

    Re: challenging rate problem

    I agree that it's difficult to see if they're both moving. But if you use the standard trick of letting John stand still the whole time and Tom run at the rate of T-J, where T is Tom's speed and J is John's, then Tom will complete a lap at this new speed in 16 minutes, giving 16(T-J)=640.

    You also know that if you combine their speeds, they do a lap in 4 minutes, so 4(T+J)=640.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,174
    Thanks
    74

    Re: challenging rate problem

    Quote Originally Posted by LoblawsLawBlog View Post
    But if you use the standard trick...
    Agree LLB; however, I feel it is better for someone to first understand...then use shortcuts...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Challenging trigonometry problem
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: February 27th 2011, 01:07 PM
  2. Challenging Problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 22nd 2009, 12:18 PM
  3. Challenging Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: August 15th 2009, 10:53 AM
  4. Challenging problem
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: July 13th 2009, 06:49 AM
  5. Challenging Ladder Problem
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 17th 2008, 07:03 PM

Search Tags


/mathhelpforum @mathhelpforum