SOLUTION: How many ounces of a solution that is 90% alcohol need to be mixed with 5 ounces of a solution that is 50% alcohol in order to obtain a solution that is 80% alcohol?
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Question 568850: How many ounces of a solution that is 90% alcohol need to be mixed with 5 ounces of a solution that is 50% alcohol in order to obtain a solution that is 80% alcohol?
I have tried a formula: 5(.50) + x(.90) = .80, but the answer yields a negative amount of ounces which is not possible. Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = ounces of 90% solution needed = ounces of alcohol in 90% solution ounces of alcohol in 50% solution
given:
15 ounces of 90% solution are needed
check answer:
OK
