<PRE><B>Question 376643</B>
How much water should be added to one gallon of pure antifreeze to obtain a solution that is 60% antifreeze? 
:
Let x = amt of water to be added, water is 0% antifreeze
:
A simple equation
1 + 0(x) = .60(x + 1)
1 = .6x + .6
1 - .6 = .6x
.4 = .6x
x = {{{.4/.6}}}
x = .67 gal of water to be added
:
:
Check by finding the amt of antifreeze before and after, should be the same
(only the percent changes)
1 = .06(1+.67)
1 = .06(1.67
1 = 1; confirms our solution 
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