Question 976855
An automobile radiator holds 16 liters of fluid.
 There is currently a mixture in the radiator that is 80% antifreeze and 20% water.
 How much of the mixture should be drained and reduced by pure antifreeze so that the resulting mixture is 90 % antifreeze?
:
let x = amt to be drained, and the amt of pure antifreeze to be added
:
.80(16-x) + x = .90(16)
12.8 - .8x + x = 14.4
-.8x + x = 14.4 - 12.8
.2x = 1.6
x = 1.6/.2
x = 8 liters to be drained and 8 liters of pure antifreeze to be added.