SOLUTION: How much 70% alcohol solution must be mixed with a 30% alcohol solution to produce 60 liters of a 40% alcohol solution?

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Question 280582: How much 70% alcohol solution must be mixed with a 30% alcohol solution to produce 60 liters of a 40% alcohol solution?
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
With solution problems like this one, you need to keep track of the 'pure' stuff involved.
The desired result is 60 liters of a 40% alcohol solution, which will have .4(60) = 24 liters of pure alcohol and 60-24 = 36 liters of water.
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The two available solutions to be used in the mixture are a 30% solution and a 70% solution
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x = volume of 30% solution
y = volume of 70% solution
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But we don't need to use two variables because we know that if we have 'x' liters of the 30% solution, then by definition we have 60-x liters of the 70% solution.
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Setting up the equation to solve the problem...
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multiply through by 10 to get rid of the decimals
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collecting and simplifying
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subtract 420 from both sides
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divide both sides by -4
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So we have 45 liters of 30% solution.
Recall we have 60 liters in total, so we can calculate the volume of 70% solution as:
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Thus our tentative answer is 45 liters of 30% solution + 15 liters of 70% solution will result in 60 liters of 40% solution.
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Of course, we always check our work!
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??
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Yes!
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Answer:
15 liters of a 70% solution must be mixed with 45 liters of a 30% solution to produce 60 liters of a 40% solution.
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Done
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