SOLUTION: A radiator contains 6 liters of a 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution?
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Question 78638: A radiator contains 6 liters of a 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution? Answer by checkley75(3666) (Show Source):
You can put this solution on YOUR website! .25(6-x)+x=.33*6
1.5-.25x+x=1.98
.75x=1.98-1.5
.75x=.48
x=.48/.75
x=.64 gallons of pure alcohol is needed to produce a 33% mixture.
proof
.25(6-.64)+.64=.33*6
.25*5.36+.64=1.98
1.34+.64=1.98
1.98=1.98
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