SOLUTION: In the lab there is a vat of a solution of 20% acid and another vat of solution of 50% acid. How much of the 50% acid solution should be mixed with the 20% solution to have 12 gal

Question 132225: In the lab there is a vat of a solution of 20% acid and another vat of solution of 50% acid. How much of the 50% acid solution should be mixed with the 20% solution to have 12 gallons of a 30% acid solution?
This was a test question the answer was 4 gallons.
No matter what I try I can't get this answer.
I've tried
.20x+.30(12)=.50(x+12)
.20x+.50(x+12)=.30(12)
.30(12) +.50(x-12)=.20x
(.20x+.50y)/12=30
Found 2 solutions by Earlsdon, stanbon:
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Well, congratulations for giving it a try!
Here's how to set it up:
Let x = the number gallons of 20% acid solution, as you did, so the amount of acid you would have is just 20% of x or 0.2x
Now you want to add (12-x) gallons of 50% acid solution (0.5(12-x) to make a total of 12 gallons of 30% acid solution (0.3(12)), so...
0.2x+0.5(12-x) = 0.3(12) Simplify and solve for x.
0.2x+6-0.5x = 3.6 Combine like-terms.
-0.3x+6 = 3.6 Subtract 6 from both sides.
-0.3x = -2.4 Finally, divide both sides by -0.3
x = 8 gallons of the 20% acid solution.
12-x = 12-8 = 4 gallons of the 50% acid solution.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
In the lab there is a vat of a solution of 20% acid and another vat of solution of 50% acid. How much of the 50% acid solution should be mixed with the 20% solution to have 12 gallons of a 30% acid solution?
---------------------------
50% solution DATA added:
Amount = "x" gallons ; Amt acid is 0.5x gallons
---------------------------
20% solution DATA:
Amount = "12-x" gallons; Amt. acid is 0.20(12-x) = 2.4-0.2x gallons
-----------------------
30% Mixture DATA:
Amount = 12 gallons: Amt acid = 0.30*12 = 3.6 gallons
-------------------------
EQUATION:
acid + acid = acid
0.5x + 2.4-0.2x = 3.6
0.3x = 1.2
x = 4 gallons ( amount of 50% solution in the mixture)
=========================
Cheers,
Stan H.
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