SOLUTION: Two acid solutions are to be mixed together. Solution A is 30% acid by volume. Solution B is 70% acid by volume. How much of solution A is needed to mix with solution B to ma

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Question 706648: Two acid solutions are to be mixed together.
Solution A is 30% acid by volume.
Solution B is 70% acid by volume.
How much of solution A is needed to mix with solution B to make an 800 mL mixture that is 54% acid by volume? Answer to the nearest millilitre.
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Your question is well organized.

a = mls. of solution A
b = mls. of solution B

System to solve:
a%2Bb=800
%2830%2Aa%2B70%2Ab%29%2F800=54

Working first through the rational equation,
%28%2830%2Aa%2B70%2Ab%29%2F800%29%2A800=54%2A800
30%2Aa%2B70%2Ab=54%2A800,
Next, divide both sides by 10 to get in this case integer coefficients.
The rational equation becomes
3a%2B7b=80%2A54.
Next we can work with the volume equation and the transformed equation that we just found:
Try subtracting 3 times the volume equation from this new form of what was the rational percentage equation, and you will easily find b. Find then a as a=800-b.

(What I said is to do this:
3a+7b-(3a+3b)=80*54-3*800,
From that find 7b-3b=1920,
and you see what to do.)