SOLUTION: A Chemist has a solution that is 5% alcohol and one that is 20% alcohol. How many liters of each should be mixed together to make 100 liters of 15% solition?

Question 220179: A Chemist has a solution that is 5% alcohol and one that is 20% alcohol. How many liters of each should be mixed together to make 100 liters of 15% solition?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 5% solution needed
Then 100-x=amount of 20% solution needed
Now we know that the amount of pure alcohol in the 5% solution (0.05x) plus the amount of pure alcohol in the 20% solution (0.20(100-x)) has to equal the amount of pure alcohol in the final mixture(0.15*100), so our equation to solve is:
0.05x+0.20(100-x)=0.15*100 get rid of parens
0.05x+20-0.20x=15 subtract 20 from each side
0.05x+20-20-0.20x=15-20 collect like terms
-0.15x=-5 divide each side by -0.15
x=33 1/3 liters----amount of 5% solution needed
100-x=100-33 1/3=66 2/3 liters amount of 20% solution needed
CK
(33 1/3)*(5/100)+(20/100)(66 2/3)=0.15*100
(100/3)(5/100)+(20/100)(200/3)=15
5/3 +40/3=15
45/3=15
15=15
Hope this helps---ptaylor
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