SOLUTION: If a home brewer has 3500mls of distilled alcohol at 80% proof how much water would he need to add to it to reduce it down to 50% proof? I've solved this part. here are my values:

Question 23160: If a home brewer has 3500mls of distilled alcohol at 80% proof how much water would he need to add to it to reduce it down to 50% proof?
I've solved this part. here are my values:
a=distilled alchohol in mls
b=proof in %
c=distilled alchohol+h2o in mls
d=h20 in mls
x=distilled alchohol in mils*proof in %
here are the steps i've used to get the h2o value
1. ab=x
2. 2x=c
3. c-a=d
my question to you math tutors is:
How can I simplify these equations into one unified equation that will solve any problem?
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I'm not sure what you did in that. It looks more complicated to me than it needs to be. Let me start over.

Let x = amount of water to be added, and let's calculate the amount of alcohol in the mixture before and after the water is added. Since you are only adding water, the amount of alcohol before equals the amount of alcohol after, and this is basis for the equation.

Amount of alcohol before = .80* 3500
Add x liters of water = No Alcohol
Amount of alcohol after = .50*(3500 + x)

Equation:
.80*3500 = .50(3500+x)
2800 = 1750+.50x
2800-1750 = .50x
1050 = .50x

Divide both sides by .50:

Add 2100 ml of water.

Check:
Beginning alcohol= .80(3500) = 2800 ml.
After water added, alcohol = .50( 3500 + 2100)= .50(5600) = 2800 ml.

R^2 at SCC
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