Results 1 to 8 of 8
Like Tree3Thanks
  • 2 Post By Prove It
  • 1 Post By ConfusedStudent123

Math Help - Algebra Word Problem

  1. #1
    Newbie
    Joined
    Jul 2013
    From
    Australia
    Posts
    3
    Thanks
    1

    Algebra Word Problem

    Hi guys! I'm kind of struggling to put this into algebraic terms, I know the answer but need to show working and just can not get my head around it, if anyone would be so kind and help me out by giving the working and an explanation as to how you got there it would be greatly appreciated. Here is the question:

    Two walkers are 15 km apart and they start walking toward each other. One walks twice as fast as other. If they meet 2 hours later at what average speed are they each walking?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Algebra Word Problem

    What have you tried to do so far?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2013
    From
    Australia
    Posts
    3
    Thanks
    1

    Re: Algebra Word Problem

    Well I've tried to make equations with what I am given and use the v = d/t formula, where one walker's speed is twice that of the other and tried solving simultaneously but it just doesn't seem to make sense.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Algebra Word Problem

    That's a good start.

    I would draw a line which can be our 15km distance. Mark in a point (it doesn't have to be accurate) where the two people will meet. The distance from the left end to that point we can call x. What would be the other distance then?
    Thanks from MarkFL and HallsofIvy
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2013
    From
    Australia
    Posts
    3
    Thanks
    1

    Re: Algebra Word Problem

    Ahhh yes! I see what you did there. Thank you for the help prove it, I look forward to our next our encounter!
    Thanks from MarkFL
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Nov 2007
    From
    Trumbull Ct
    Posts
    888
    Thanks
    25

    Re: Algebra Word Problem

    Hello ConfusedStudent123,

    Rate of closure of two bodies approaching equals closure distance /time
    15/2 = 7.5 km/hr
    rate of A =x B=2x
    3x =7.5 x= 2.5 km/hr 2x = 5.0 km/hr
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jul 2013
    From
    Norway
    Posts
    8

    Re: Algebra Word Problem

    I would do it very simply.
    The slow walker would be X and because the faster walker is twice as fast he would be 2X. Add them together you get 3X. 15/3=5. X=5 2X=10.
    Devide those with 2 and you should get the avrage speed in km/h.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Algebra Word Problem

    Quote Originally Posted by Pusekatten View Post
    I would do it very simply.
    The slow walker would be X and because the faster walker is twice as fast he would be 2X. Add them together you get 3X. 15/3=5. X=5 2X=10.
    So you set 3X equal to 15? Why? 3X is a speed (the relative speed with which one is approaching the other) and 15 is a distance. They can't be equal!

    Devide those with 2 and you should get the avrage speed in km/h.
    Divide what with 2? And which "average speed"? You understand that they are asking for two average speeds don't you? If you solve "3X= 15, you get X= 5. But you just said "the slow walker would be X" and, in fact, "5 km/hr" is the speed of the faster walker!

    Confusedstudent123, Pusekatten ("pussycat"?) does have the basic idea. If you call the speed of the slower walker "X", in km/hr, then the speed of the other, who is walking twice as fast, is 2X. Each is approaching the other at speed 3X. So this is exactly the same as if one person stood still and the other walked toward him at speed 3X. (In very technical terms, this would be "relative" to that person's "frame of reference".)

    So this is exactly the same as if one person walked 15 km in 2 hours. That person would be walking at an average of 15/2= 7.5 km/hr. But that is equal to 3X so we have 3X= 7.5, X= 7.5/3= 2.5 km/hr. That is the speed of the slower person. The other person, walking twice as fast, would be walking at 2(2.5)= 5 km/hr.

    As a check, the slower walker, at 2.5 km/hr, will have walked, in 2 hours, 2(2.5)= 5 km. The faster walker will have walked 2(5)= 10 km.
    Yes, those add to 15 km so they will have met at the end of those 2 hours.
    Last edited by HallsofIvy; July 22nd 2013 at 11:56 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with Algebra Word Problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 19th 2012, 12:01 PM
  2. Algebra 2 Word Problem #2
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 2nd 2009, 02:48 PM
  3. Algebra 2 word problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 2nd 2009, 02:37 PM
  4. algebra word problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: December 13th 2008, 08:48 AM
  5. Algebra Word Problem-HELP!
    Posted in the Algebra Forum
    Replies: 8
    Last Post: July 25th 2008, 01:57 AM

Search Tags


/mathhelpforum @mathhelpforum