SOLUTION: I've been trying to figure out how to do this problem:
What fraction of 10% alcohol should be mixed with 70% alcohol to obtain 50% alcohol?
I've been trying to use the alligatio
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Question 270456: I've been trying to figure out how to do this problem:
What fraction of 10% alcohol should be mixed with 70% alcohol to obtain 50% alcohol?
I've been trying to use the alligations method, but not sure if that is the way to go about it...could you clarify the equation formula for me please?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
.10x+.70y=.50(x+y)
.10x+.70y=.50x+.50y
.10x-.50x-.50y+.70y=0
.40x+.20y=0
This tells us that twice as much of the 70% solution (y) is used as the 10% solution (x).
Proof:
.10*1+.70*2=.50(1+2)
.1+1.4=.5*3
1.5=1.5
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