SOLUTION: an automobile radiator holds 6 liters of fluid. there is currently a mixture in the radiator that is 80% antifreeze and 20% water. how much of this mixture should be drained and r

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: an automobile radiator holds 6 liters of fluid. there is currently a mixture in the radiator that is 80% antifreeze and 20% water. how much of this mixture should be drained and r      Log On

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Question 213393: an automobile radiator holds 6 liters of fluid. there is currently a mixture in the radiator that is 80% antifreeze and 20% water. how much of this mixture should be drained and replaced by pure anitfreezes so that the resulting mixture is 90% antifreeze?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
the best way (for me) to solve this is to
express an equation in words that sum up the
whole situation.
The 1st thing to realize is that you start out with
6 liters of fluid and you end up with 6 liters
of fluid
Here it is in words:
((original amount of antifreeze) - (amount of antifreeze drained out) + (amount of antifreeze put back in))
[all divided by]
(total amount of fluid I end up with) = 90%
-------------------------------------------
Let x= liters of fluid I drain out and replace with pure antifreeze
Original amount of antifreeze = .8%2A6+=+4.8 liters
Amount of antfreeze drained out = .8x
Amount of antifreeze put back in = x
------------------------------------------
Now I can write the equation
%284.8+-+.8x+%2B+x%29%2F6+=+.9
4.8+-+.8x+%2B+x+=+5.4
.2x+=+5.4+-+4.8
.2x+=+.6
x+=+3
3 liters, or 1/2 the fluid in the radiator must
be drained and replaced with pure antifreeze to get
90% antifreeze
check answer:
There must be 10% water in the final solution
I start with .2%2A6+=+1.2 liters water
I drain out .2x+=+.2%2A3 or .6 liters water
I end up with 1.2+-+.6+=+.6 liters water
.6+=+.1%2A6
So, I do end up with 10% water