<PRE><B>Question 199976</B>
How much water should be added to 1 gallon of pure antifreeze to obtain
 a solution that is 60% antifreeze?
:
Let x = amt of water required to accomplish this
:
1.0(1) = .6(x+1) 
1 = .6x + .6
1 - .6 = .6x
.4 = .6x
x = {{{.4/.6}}}
x = {{{2/3}}} gallon of water required
:
:
Check solution in the original equation (use .67 for 2/3 gal)
1 = .6(.67+1)
1 = .6(1.67)
1 = 1.00</PRE>