Question 396161
A radiatior contains 6 liters of a 20% antifreeze solution. How much should be drained and replaced with pure antifeeze to produce a 50% antifreeze solution?
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Let x = amount (liters) removed and replaced by pure antifreeze
then
.20(6-x) + x = .50(6)
1.2-.2x + x = 3
1.2 + .8x = 3
.8x = 1.8
x = 1.8/.8
x = 2.25 liters