SOLUTION: A chemist needs to mix a 75% acid solution with a 50% acid solution to obtain 10 milliliters of a 60% acid solution. How many milliliters of each of the solutions must be used?

Question 531088: A chemist needs to mix a 75% acid solution with a 50% acid solution to obtain 10 milliliters of a 60% acid solution. How many milliliters of each of the solutions must be used?
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
A KMST (chemist) is the right person to answer this question.
We have to assume that volumes are additive in this case, meaning that the volume after mixing will equal the sum of the volumes before mixing. (Not always true).
We also have to assume that the concentrations were given in percentage weight in volume (grams of acid per milliliter of solution). (Not usually the case, but done in some cases).
I want to assume that it is the same acid, because we do not like to add apples and oranges, and we know that millions of acids exist.
Let s be the volume (in milliliters) of 75% solution (strong solution) to be used.
Let w be the volume (in milliliters) of 50% solution (weak solution) to be used.
The amount of acid in each volume used will be 0.75s and 0.50w.
The amount of acid in the final mixture has to be
That leads us to the equations
and

If studying systems of linear equations, solve as you are being taught to do.
If not there yet, change to
substitute in the other equation to get
,
and
and
Then find

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