# Word Problem: Computing age comparing quantities

• Jul 9th 2012, 02:58 PM
allyourbass2212
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 • Jul 9th 2012, 03:15 PM Plato Re: Word Problem: Computing age comparing quantities Quote: Originally Posted by allyourbass2212 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?

$m+j+a=3j+j+j-2=43$
• Jul 9th 2012, 03:23 PM
allyourbass2212
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.
• Jul 9th 2012, 03:29 PM
Plato
Re: Word Problem: Computing age comparing quantities
Quote:

Originally Posted by allyourbass2212
. 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 $x + (\frac{x}{3})+ (\frac{x}{3} -2) = 43$
• Jul 9th 2012, 03:32 PM
skeeter
Re: Word Problem: Computing age comparing quantities
Quote:

Originally Posted by allyourbass2212
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