SOLUTION: An automobile radiator contains 16 liters of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so t

Question 225271: An automobile radiator contains 16 liters of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze?
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = liters of mixture to be drained
and replaced with pure antifreeze.
In words:
(liters of antifreeze I end up with)/(final total liters in radiator) = 50%
Notice that I start with 16 liters in the radiator
and I end up with 16 liters in the radiator, so I can write
(liters of antifreeze I end up with)/16 = 50%
16 liters of 30% antifreeze has liters antifreeze
I drain out liters, and in doing so, I remove .3x liters
antifreeze
So far, I've got liters antifreeze in radiator
Now I pour in liters of antifreeze
I end up with liters antifreeze
My equation is then

liters
4.571 liters need to be drained out and
replaced with pure antifreeze
check answer
If I replace with a 30% mixture again, I'll be back
to 30% in the whole radiator
liters antifreeze

liters antifreeze in rest of radiator

OK

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