SOLUTION: A chemist needs to combine two solutions of a certain acid. How many liters of a 30% acid solution should be added to 40 liters of a 12% acid solution to obtain a final mixture th

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Question 158772This question is from textbook Begining Algebra: Early Graphing
: A chemist needs to combine two solutions of a certain acid. How many liters of a 30% acid solution should be added to 40 liters of a 12% acid solution to obtain a final mixture that is 20% acid? This question is from textbook Begining Algebra: Early Graphing

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!
Let x=amount of 30% acid solution needed
Now we know that the amount of pure acid in the 30% solution (0.30x) plus the amount of pure acid in the 40 liters of 12% solution (0.12*40) has to equal the amount of pure acid in the solution after the mixture takes place (0.20(40+x)). So our equation to solve is:
0.30x+0.12*40=0.20(40+x) get rid of parens
0.30x+4.8=8+0.20x subtract 4.8 and also 0.20x from both sides
0.30x-0.20x+4.8-4.8=8-4.8+0.20x-0.20x collect like terms
0.10x=3.2 divide each side by 0.10
x=32 gal------------------------------amount of 30% solution needed
CK
0.30*32+0.12*40=0.20*72
9.6+4.8=14.4
14.4=14.4
Hope this helps----ptaylor