SOLUTION: HOW MANY LITERS OF A 10% ALCOHOL SOLUTION MUST BE MIXED WITH 40 L OF A 50% SOLUTION TO GET A 40% SOLUTION?

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Question 212320: HOW MANY LITERS OF A 10% ALCOHOL SOLUTION MUST BE MIXED WITH 40 L OF A 50% SOLUTION TO GET A 40% SOLUTION?
Answer by HyperBrain(694) About Me  (Show Source):
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Let x=the amount of 10% alcohol solution required
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x L + 40 L = (x+40) L of mixture
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If we have x L of 10% alcohol solution, we have 0.10x L pure alcohol
If we have 40 L of 50% alcohol solution, we have 0.5(40)=20 L pure alcohol
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So, we have the equation:
0.40(x+40)=0.10x+20
40(x+40)=10x+2000
4(x+40)=x+200
4x+160=x+200
3x=40
x=40/3 L