SOLUTION: Beer A is 6% alcohol, Beer B is 2% alcohol.
how many liters of beer A should be mixed in order to get 50 liters of mixture that is 3.2% alcohol.
solution A + B = 50 liter
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Question 383323: Beer A is 6% alcohol, Beer B is 2% alcohol.
how many liters of beer A should be mixed in order to get 50 liters of mixture that is 3.2% alcohol.
solution A + B = 50 liter
6%+2% = 3.2% alcohol
.06 + .02= 3.2
I am stuck on this calculation, need to get one of the variables alone...like A but am not sure where to go from this point...
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
Let A be the amount of beer A, B the amount of beer B.
1.
.
.
2.
Now you have two equations in two unknowns.
Subtract eq. 1 from eq. 2 to eliminate B and solve for A.
Then go back to either equation and solve for A.
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