SOLUTION: 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 re

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?
Found 2 solutions by FrankM, ankor@dixie-net.com:
Answer by FrankM(1040) (Show Source): You can put this solution on YOUR website!
90% of 16l is 14.4l antifreeze and 1.6l water.
What we have now is 20% water. Since we can only have 1.6l water, 8l of the old mix is all we can keep. Dump out 8l and add pure antifreeze to get a 90% mix.

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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.
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