<PRE><B>Question 13262</B>
Let x = amount of water to be added with 0% antifreeze
1 = amount of pure antifreeze at 100%
x+1 = amount of the mixture at 60% antifreeze


{{{0* x + 1.00 * 1 = .60(x+1)}}}
1 = .60x + .60


Subtract .60 from each side:
1-.60 = .60x
.40 = .60x


Divide both sides by .60:

{{{.40/.60 = (.60x)/.60 }}}

{{{x = .4/.6 = 4/6 = 2/3}}} gallon.


Check:  Total antifreeze = 1 gallon in total mixture of 1 + 2/3 gallon = 5/3 gallons.  1 gallon divided by 5/3 equals 3/5 which is 60%.  It checks.


R^2 at SCC 
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