SOLUTION: It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 70 liters of 20% solution. How many liters of this should be drained and re

Question 121689This question is from textbook College Algebra
: It is necessary to have a 40% antifreeze solution in the radiator of a certain car. The radiator now has 70 liters of 20% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength? This question is from textbook College Algebra

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = the amount of 20% solution that I need to
drain off.
Then will also be the amount of pure antifreeze
that I add afterwards.
My basic equation in words is:
(% antifreeze solution) = (amount of antifreeze) / (antifreeze + water)
My (antifreeze + water) does not change. I started with l
and I end up with l.
I'm looking for a 40% solution
.4 = (amount of antifreeze)/70
I start off with l of 20% solution.
That means that it contains
l of antifreeze
I first drained off amount of 20% solution.
That means I drained off amount of antifreeze
The I add amount of antifreeze

l answer
I'd like to check this
What did I start with?
l antifreeze and l water for a l solution
I drain off l of this 20% solution
I drain off l antifreeze
and l water
Now I've got l antifreeze
and l water
The last step: I want to add l of antifreeze
Now I've got l of antifreeze
and still l of water
going back to:
(% antifreeze solution) = (amount of antifreeze) / (antifreeze + water)
(% antifreeze solution) =
(% antifreeze solution) =
(% antifreeze solution) = 40%
OK
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