Question 1049640
A radiator contains 8 quarts of a mixture of water and antifreeze that is 40% antifreeze,
 how much of the mixture should be drained and replaced by pure antifreeze so that the resultant mixture contain 60% antifreeze?
:
let x = of pure antifreeze and also = the amt of mixture to be removed
:
.4(8-x) + x = .60(8)
3.2 - .4x + x = 4.8
.6x = 4.8 - 3.2
.4x = 1.6
x = 1.6/.4
x = 4 qts removed and 4 qts of pure antifreeze added
:
In terms of percent, 50% removed and 4 qts added