SOLUTION: Please help to solve One solution was 10% alcohol and the other was 30% alcohol. HOw much of each should be used to get 200 gallons of a solution that is 17% alcohol?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Please help to solve One solution was 10% alcohol and the other was 30% alcohol. HOw much of each should be used to get 200 gallons of a solution that is 17% alcohol?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 68592This question is from textbook An Incremental Development
: Please help to solve
One solution was 10% alcohol and the other was 30% alcohol. HOw much of each should be used to get 200 gallons of a solution that is 17% alcohol? This question is from textbook An Incremental Development

Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

Let x= amount of 30% alcohol
Then 200-x=amount of 10% alcohol
Now we know that the amount of pure alcohol in the 10% solution 0.10(200-x) plus the amount of pure alcohol in the 30% solution 0.30x equals the amount of pure alcohol in the final mixture 0.17(200). So our equation to solve is:
0.10(200-x)+0.30x=0.17(200) clear parens
20-0.10x+0.30x=34 subtract 20 from both sides
+0.20x=14
x=70 gal of 30% alcohol
200-x=200-70=130 gal of 10% alcohol
CK
70(0.3)+130(0.1)=200(.17)
21+13=34
34=34

Hope this helps --------ptaylor