SOLUTION: How many liters of a 30% antifreeze solution and 60% antifreeze solution must be mixed to make 10 liters of a 39% antifreeze solution

Question 923292: How many liters of a 30% antifreeze solution and 60% antifreeze solution must be mixed to make 10 liters of a 39% antifreeze solution
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=number of liters of the 30 % solution needed
Then 10-x=number of liters of the 60% solution needed
Now we know that the amount of pure antifreeze that exists before the mixture takes place(0.30X+0.60(10-x)) has to equal the amount of pure antifreeze that exists after the mixture takes place (0.39*10). Soooo our equation to solve is:
0.30x+0.60(10-x)=0.39*10 simplify
0.30x+6-0.60x=3.9 collect like terms
-0.30x=-6+3.9=-2.1
x=7 liters------amount of 30% solution needed
10-x=10-7=3 liters amount of 60% solution needed
CK
0.30*7+0.60*3=0.39*10
2.1+1.8=3.9
3.9=3.9
Hope this helps---ptaylor
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