SOLUTION: how many liters of a 10% acid solution must be mixed with a 50% acid solution in order to obtain 20 liters of a 30% solution?

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Question 69112: how many liters of a 10% acid solution must be mixed with a 50% acid solution in order to obtain 20 liters of a 30% solution?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
.1(20-x)+.5x=.2*20
2-.1x+.5x=4
.4x=2
x=2/.4
x=5 liters of 50% acid solution
20-5=15 liters of 10% acid solution
proof
.1*15+.5*5=4
1.5+2.5=4
4=4
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MY APPOLOGIES: I MISTAKENLY USED 20% OF 20 LITERS INSTEAD OF 30% OF 20 LITERS:
THUS THE EQUATION SHOULD HAVE BEEN
.1(20-X)+.5X=.3*20
2-.1X+.5X=6
.4X=6-2
.4X=4
X=4/.4
X=10 LITERS OF 50% IS REQUIRED TO MAKE 20 LITERS OF 30%.
I MUST HAVE BEEN HALF ASLEEP WHEN I DID THIS PROBLEM LAST NIGHT. AGAIN I'M SORRY FOR THE MISTAKE.
HOPE THIS WILL NOT DISCOURAGE YOU FROM USING ALGEBRA.COM IN THE FUTURE TO HELP WITH YOUR MATH DIFFICULTIES.