1. ## Ratio Proportion

I'm stuck up with this too

The ages of Katheline and Gwen are in the ratio 2 : 3. After 12 years, their ages will be in the ratio 11 : 15. The age of Gwen is-
a) 32 years
b) 42 years
c) 48 years
d) 56 years

Any help will be appreciated.

Thanks a Lot!

2. ## Re: Ratio Proportion

Originally Posted by MoniMini
I'm stuck up with this too

Any help will be appreciated.

Thanks a Lot!
Let x denote the age of Katheline and y the age of Gwen.

Then you know:

$\frac xy = \frac23$

$\frac{x+12}{y+12} = \frac{11}{15}$

This is a system of simultaneous equations which you can solve easily.

3. ## Re: Ratio Proportion

Ok.......
Initially I had come to this conclusion but
I'm having some troubles after this equation.
How to find the values of x and y?

4. ## Re: Ratio Proportion

Usually in solving a system of simultaneous equations, you wind up manipulating one of them so you can express on variable in terms of the other. That allows you to substitute that variable into the other equation.

Let's take this instance as an example: let's call the top equation Equation 1, and the lower one Equation 2.

Equation 1 can be re-worked so we can express x in terms of y:

x/y = 2/3

3x = 2y

So we can then figure out

x = 2y/3

Knowing that, we can then substitute this value for x into Equation 2. When you do that, what do you get then?

5. ## Re: Ratio Proportion [SOLVED]

Oooh! I got it now!
$\frac xy = \frac23$
and
$\frac{x+12}{y+12} = \frac{11}{15}$

then

$\15 (x + 12) = 11 (y + 12)$

$= 15x + 180 = 11y + 132$

$= 180 - 132 = 11y - 15x$

$= 48 = 11y - 15x$

and if

$\frac xy = \frac23$

then

$x = \frac {2y} 3$

then

$15x = \frac {15} 1 \!\cdot\! \frac {2y} 3$
$= \frac {30y}3$

then we go back to

$= 48 = 11y - 15x$

so that becomes

$48 = 11y - \frac {30y}3$
$= 48 = 11y - 10y$
$= 48 = y$

So, if $x$ is Katheline and $y$ is Gwen, then
Gwen is 48 years old.

Whoa!!
Thnx a lot @earboth and @mathbyte!

~MoniMini