SOLUTION: Some 80% acid solutions is to be mixed with 35% acid solution to make 300 liters of 50% solution. How much of each acid should be used?

Question 873567: Some 80% acid solutions is to be mixed with 35% acid solution to make 300 liters of 50% solution. How much of each acid should be used?
Found 2 solutions by josgarithmetic, ghostrecog:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Here is a lesson on this general kind of problem because this type of question is asked very frequently. The solution is also shown in the lesson.

http://www.algebra.com/tutors/Mixtures%3A-All-in-Symbols.lesson?content_action=show_dev
Answer by ghostrecog(4) (Show Source): You can put this solution on YOUR website!
Solution:

http://www.algebra.com/cgi-bin/display-illustration.mpl?tutor=ghostrecog&illustration_name=solution.bmp&size=thumbnail
Refer from the above url, consists of table which explains the below equation.

Then the Equation as follows
0.80x+0.35(300-x)=300(.50)
0.80x+105-0.35x=150
0.45x=45
x=100 of 80% of acid solution
300-x=200 of 35% of Acid solution. When mixed together we get 300 liters of solution.
check:
0.80(100) =80 liters
0.35(300-x)=70 liters
----------------------
300(.50) =150 liters
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