SOLUTION: Solve: How much 2% alcohol must be mixed with 10% alcohol to get 2000ml of 8% alcohol? Could you please show me step by step how you get the answer to this problem please? It's

Question 399915: Solve: How much 2% alcohol must be mixed with 10% alcohol to get 2000ml of 8% alcohol?
Could you please show me step by step how you get the answer to this problem please? It's an applied problem. I just don't know how you solve this problem.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 2% alcohol needed
Then 2000-x=amount of 10% alcohol needed
Now we know that the amount of pure alcohol that we have before the mixture takes place has to equal the amount of pure alcohol that we have after the mixture takes place.
Amount of pure alcohol before the mixture takes place: 0.02x+0.10(2000-x)
Amount of pure alcohol after the mixture takes place: 0.08*2000
So, our equation to solve is:
0.02x+0.10(2000-x)=0.08*2000 simplify and get rid of parens
0.02x+200-0.10x=160 subtract 200 from each side and collect like terms
-0.08x=-40 divide each by -0.08
x=500ml-------------amount of 2% alcohol needed
2000-500=1500ml---------------amount of 10% alcohol needed
CK
0.02*500+0.10*1500=0.08*2000
10+150=160
160=160

Hope this helps---ptaylor

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