SOLUTION: How many liters of a mixture containing 70% alcohol should be added to a mixture containing 20% alcohol to obtain 16L of a mixture containing 50% alcohol?
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Question 274164: How many liters of a mixture containing 70% alcohol should be added to a mixture containing 20% alcohol to obtain 16L of a mixture containing 50% alcohol? Answer by fractalier(6550) (Show Source):
You can put this solution on YOUR website! Let the amount of 70% alcohol be x.
Since we need a total of 16 L, the amount of 20% alcohol used must be 16 - x.
The setup looks like this
.70x + .20(16-x) = .50(16)
Multiply everything by ten to clear decimals
7x + 2(16-x) = 5(16)
7x + 32 - 2x = 80
5x + 32 = 80
5x = 48
x = 9.6 L
