SOLUTION: How much water should be added to 30 gallons of a solution that is 70% antifreeze in order to get a mixture that is 60% antifreeze. How do I go about setting up this equation?

Question 66467: How much water should be added to 30 gallons of a solution that is 70% antifreeze in order to get a mixture that is 60% antifreeze.
How do I go about setting up this equation? Please help!
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
How much water should be added to 30 gallons of a solution that is 70% antifreeze in order to get a mixture that is 60% antifreeze.
How do I go about setting up this equation? Please help!
Let x=amount of water to be added
Now we know that the amount of pure antifreeze in the 70% solution (30)(.70)plus the amount of pure antifreeze that is added(0)(we are adding pure water) must equal the amount of pure antifreeze in the final solution (30+x)(.60)
So our equation to solve is:
(1) (30)(.70)=(30+x)(.60)
That's how you set it up
Now we can also solve this problem by dealing with the water:
We will still let x=amount of water to be added
Now we know that the amount of pure water in the original solution (30)(.30) plus the amount of pure water added (x) equals the amount of pure water in the final mixture (30+x)(.40). So our equation to solve is:

(2) (30)(.30)+x=(30+x)(.40)
Now this should give you the same answer for x so we'll solve it and plug the answer into (1).
(30)(.30)+x=(30+x)(.40)
9+x=12+.4x
.6x=3
x=5 gals of pure water that needs to be added
Ck
substitute into (1)
(30)(.70)=(30+x)(.60)
21=35(.6)
21=21

Hope this helps-----ptaylor

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