SOLUTION: How many liters of a 10% alcohol solution must be mixed with 40 liters of a 50% solution to get a 40% solution? I got the answer 13.3 L, but I'm not sure that it's right.

Question 507710: How many liters of a 10% alcohol solution must be mixed with 40 liters of a 50% solution to get a 40% solution?
I got the answer 13.3 L, but I'm not sure that it's right.
Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
With mixture problems, you need to pay attention to how much 'pure' stuff you have and how much you need.
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40 liters of 50% solution means you have 20 liters of 'pure' alcohol
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x = liters of 10% solution is to be added until you have a final solution that is 40% alcohol.
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.5(40) + .1(x) = .4(40+x)
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multiply by 10 to remove decimals
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5(40) + x = 4(40+x)
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200 +x = 160 + 4x
40 = 3x
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x = 40/3 = 13 1/3 liters
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Check this to be sure it is the answer.
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40/3 * .1 = 4/3 liters of alcohol
120/3 * .5 = 20 liters of alcohol
= 21 1/3 liters of alcohol.
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40 + 40/3 = 120/3 + 40/3 = 160/3
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160/3*.4 = 64/3 = 21 1/3 liters of alcohol
which is correct.
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Answer: Add 13 1/3 liters of 10% alcohol solution.
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Done.
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