Question 472759
A radiator contains 8 liters of mixture of water and antifreeze.
 If 40% of the mixture is antifreeze, how much of the mixture shall be drained
 and replaced with pure antifreeze so that the resultant mixture will contain 60% antifreeze?
:
Let x = the amt to be drained, and the amt of pure antifreeze to be added
:
.40(8-x) + 1x = .60(8)
3.2 - .4x + 1x = 4.8
-.4x + 1x = 4.8 - 3.2
.6x = 1.6
x = {{{1.6/.6}}}
x = 2{{{2/3}}} liters drained and 2{{{2/3}}} liters of pure antifreeze is added