SOLUTION: It is desired to have a 20-liter mixture of 60% alcohol. Two mixtures, one or 85% alcohol another of 35% of alcohol are to be used. How many liters of each will be required?

Question 1126457: It is desired to have a 20-liter mixture of 60% alcohol. Two mixtures, one or 85% alcohol another of 35% of alcohol are to be used. How many liters of each will be required?
Found 3 solutions by addingup, josgarithmetic, Alan3354:
Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
We want a total of 20 liters. Let the amount to be used of the 85% mixture be x. Then, the amount of the mixture at 35% will be 20 - x
0.85x + 0.35(20 - x) = 0.6(20)
0.85x + 7 - 0.35x = 12
0.5x = 5
x = 10 liters of the 85% alcohol mixture and 20-10 = 10 liters of the 35% mixture are needed to obtain a mixture of 60% alcohol

Answer by josgarithmetic(39627)   (Show Source): You can put this solution on YOUR website!
Use variable for the higher available concentration material.

x of 85%
20-x of 35%
Want 60% alcohol

.
.
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
60 is the average 0f 85 & 35
--> equal amounts.
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