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(39618) (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:
Working first through the rational equation,
,
Next, divide both sides by 10 to get in this case integer coefficients.
The rational equation becomes
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 .
(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.)
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