Word problem method needed

Hi, I am looking for a method similar to age problems or rate, dist problems for getting a proportion of a mixture.

In this example I have to mix coffee1 at $6.50/unit with cofee2 at $9.00/unit to give me $7.50/unit.

I got the answer by doing it this way;

(6.50 + 9.00 )/7.5 = x

6.5/x + 9/x = 7.5

So 6.5/x is my coffee1 component.

Is this a accepted method or how would a diagram be used here? Anything that would make life easier will be good and will help remember what to do next time.

Thanks for the help.

Re: Word problem method needed

Quote:

Originally Posted by

**SVDM**

(6.50 + 9.00 )/7.5 = x

Ehh, I don't think that works. That equation yields x = 2.066..., but we don't want x (also, what does x represent?).

A better way to solve it is let a and b be the number of units of coffee1 and coffee2, respectively. We take a weighted average:

$\displaystyle \frac{6.5a + 9b}{a+b} = 7.5$ For the LHS, the numerator represents the total cost of the coffee, the denominator represents the total # of units.

$\displaystyle 6.5a + 9b = 7.5a + 7.5b \Rightarrow 1.5b = a$. That means, for every 1 unit of b (coffee2), we need 1.5 units of a (coffee1) to stay at $7.50 a unit. We like integer ratios so I'll multiply by 2, for every 3 units of coffee1 we need 2 units of coffee2.