Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By Plato

Math Help - Word Problem: Computing age comparing quantities

  1. #1
    Member
    Joined
    May 2008
    Posts
    143

    Word Problem: Computing age comparing quantities

    Problem: Monica earns three times per hour as John, John earns $2 more per hour than Alicia. Together they earn $43 per hour. How much is each one's hourly wage?

    I attempted to compare each quantity in comparison to Monica

    Monica = x
    John = Monica earns three times per hour as John, therefore John earns 1/3 in comparison to Monica = \frac{1}{3x}
    Alicia = receives $2 less than John = = \frac{1}{3x}-2

    The equation then becomes

    x + (\frac{1}{3x})+ (\frac{1}{3x} -2) = 43

    Clear out the fractions by multiplying every item by the LCD of 3, so on so forth but I do not get the correct answer.

    The correct answer is x = 7, did I make an error along the way or is all of the work completely wrong?

    Thanks for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,389
    Thanks
    1476
    Awards
    1

    Re: Word Problem: Computing age comparing quantities

    Quote Originally Posted by allyourbass2212 View Post
    Problem: Monica earns three times per hour as John, John earns $2 more per hour than Alicia. Together they earn $43 per hour. How much is each one's hourly wage?
    Make it easy on yourself. Start with John.
    m+j+a=3j+j+j-2=43
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    143

    Re: Word Problem: Computing age comparing quantities

    Thanks Plato. Yes that is how I solved the problem after getting stuck on my initial approach. But I would still like to know why I am not having success comparing quantities using Monica.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,389
    Thanks
    1476
    Awards
    1

    Re: Word Problem: Computing age comparing quantities

    Quote Originally Posted by allyourbass2212 View Post
    . But I would still like to know why I am not having success comparing quantities using Monica.
    This x + (\frac{1}{3x})+ (\frac{1}{3x} -2) = 43 wrong.

    It should be x + (\frac{x}{3})+ (\frac{x}{3} -2) = 43
    Thanks from allyourbass2212
    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

    Re: Word Problem: Computing age comparing quantities

    Quote Originally Posted by allyourbass2212 View Post
    Thanks Plato. Yes that is how I solved the problem after getting stuck on my initial approach. But I would still like to know why I am not having success comparing quantities using Monica.
    monica, x

    john, \frac{x}{3}

    alicia, \frac{x}{3}-2

    x + \frac{x}{3} + \frac{x}{3} - 2 = 43

    x + \frac{x}{3} + \frac{x}{3} = 45

    3x + x + x = 135

    5x = 135

    x = 27, monica

    \frac{x}{3} = 9, john

    \frac{x}{3} - 2 = 7, alicia
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help computing diminishing returns problem
    Posted in the New Users Forum
    Replies: 1
    Last Post: June 20th 2012, 08:00 PM
  2. Comparing Two Quantities
    Posted in the Algebra Forum
    Replies: 8
    Last Post: February 28th 2011, 07:24 PM
  3. Replies: 2
    Last Post: September 29th 2010, 12:58 PM
  4. Replies: 2
    Last Post: July 31st 2010, 06:05 AM
  5. computing sin of an angle (difficult problem)
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: January 7th 2010, 10:08 PM

Search Tags


/mathhelpforum @mathhelpforum