Results 1 to 5 of 5

Math Help - Math word problems

  1. #1
    Newbie
    Joined
    Dec 2009
    Posts
    7

    Math word problems

    Hi, new here. I got a question here that I do not understand how to do at all. It'd be nice if someone explained how to solve it.

    It took a crew 80 minutes to row 3km upstream and back again. If the rate of the flow of the stream was 3km/h, what was the rowing rate of the crew.

    I did:

    3=(1.33)(x+3)
    3=(1.33)(x-3)

    I was going to solve by substitution but I realized that 1.33 cannot be the time for both going up and downstream.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by brownjovi View Post
    Hi, new here. I got a question here that I do not understand how to do at all. It'd be nice if someone explained how to solve it.

    It took a crew 80 minutes to row 3km upstream and back again. If the rate of the flow of the stream was 3km/h, what was the rowing rate of the crew.

    I did:

    3=(1.33)(x+3)
    3=(1.33)(x-3)

    I was going to solve by substitution but I realized that 1.33 cannot be the time for both going up and downstream.
    let r = row rate in km/hr

    t = time in hrs upstream

    \frac{4}{3} - t = time in hrs downstream

    (r-3)t = 3

    (r+3)\left(\frac{4}{3} - t\right) = 3


    from the 1st equation ... t = \frac{3}{r-3}

    sub this into the 2nd equation ...

    (r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3

    solve for r
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121
    Since we do not know the time taken to row upstream and downstream separately, let " t_1" be the time taken to row upstream and let " t_2" be the time taken to row downstream. Let "v" be their speed in still water.

    Then their speed upstream is v- \frac{4}{3} and that equals "distance divided by time: v- \frac{4}{3}= \frac{3}{t_1}.

    Their speed downstream is v+ \frac{4}{3} and so v+ \frac{4}{3}= \frac{3}{t_2}.

    Those, together with t_1+ t_2= 80 give three equations to solve for the three values t_1, t_2, and v. Of course, you are only asked for v so if you could find a way to eliminate t_1 and t_2, reducing to a single equation in v, that would be excellent.
    Last edited by HallsofIvy; December 10th 2009 at 03:14 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2009
    Posts
    7
    Quote Originally Posted by skeeter View Post

    sub this into the 2nd equation ...

    (r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3
    How would I remove the r-3 from

    (\frac{4}{3} - \frac{3}{r-3})

    ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by brownjovi View Post
    How would I remove the r-3 from

    (\frac{4}{3} - \frac{3}{r-3})

    ?
    (r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3

    (r+3)\left(\frac{4(r-3)}{3(r-3)} - \frac{9}{3(r-3)}\right) = 3

    (r+3)\left(\frac{4(r-3)-9}{3(r-3)}\right) = 3

    (r+3)\left(\frac{4r-21}{3(r-3)}\right) = 3

    \frac{(r+3)(4r-21)}{3(r-3)} = 3

    (r+3)(4r-21) = 9(r-3)

    expand both sides, combine like terms and solve the resulting quadratic equation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: March 29th 2011, 07:11 PM
  2. Funcation math word problems.. Help?!?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 1st 2010, 08:24 AM
  3. Replies: 0
    Last Post: April 27th 2010, 07:19 PM
  4. Math Word Problems: Investment
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 30th 2009, 08:20 PM
  5. Math Word Problems
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 14th 2008, 07:25 PM

Search Tags


/mathhelpforum @mathhelpforum