SOLUTION: How many liters of 10% alcohol solution and 5% alcohol solution must be mixed to get 40 liters of 8%?
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Question 1004980: How many liters of 10% alcohol solution and 5% alcohol solution must be mixed to get 40 liters of 8%?
Found 2 solutions by stanbon, ikleyn:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How many liters of 10% alcohol solution and 5% alcohol solution must be mixed to get 40 liters of 8%?
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Equation:
alcohol + alcohol = alcohol
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0.10x + 0.05(40-x) = 0.08*40
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10x + 5*40 - 5x = 8*40
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5x = 3*40
x = 24 liters (amt. of 10% solution needed)
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40-x = 16 liters (amt. of 5% solution needed)
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Cheers,
Stan H.
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Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
Similar problem was solved in the lesson Solving typical word problems on mixtures for solutions in this site (Problem 1).
Read it attentively and then solve your problem by substituting your data.
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