SOLUTION: If I have a 20 qt radiator containing a 80% antifreeze solution. How much of the solution should I drain and replace with pure water to get a solution that is 50% antifreeze?
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Question 420867: If I have a 20 qt radiator containing a 80% antifreeze solution. How much of the solution should I drain and replace with pure water to get a solution that is 50% antifreeze? Found 3 solutions by mananth, ikleyn, josgarithmetic:Answer by mananth(16949) (Show Source):
You can put this solution on YOUR website! Let 82% solution required be x
percent ---------------- quantity
82 ---------------- x
water 0% ---------------- 20-x
50.00% ---------------- 20
...
82x+0(20 -x)=50*20
82x=1000
/ 82
x=12.2 qt. of82.00%
Balance 7.8 qt of water.
7.8 qt. of 82% will have to be drained and replaced with water
You can put this solution on YOUR website! .
If I have a 20 qt radiator containing a 80% antifreeze solution, how much of the solution
should I drain and replace with pure water to get a solution that is 50% antifreeze?
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The solution in the post by @mananth is incorrect.
I came to bring a correct solution.
Let x be the volume of the original 80% antifreeze solution to partly drain
and to replace with pure water to get the 40% antifreeze solution.
After draining, we then have (20-x) quartz of the 80% antifreeze solution.
It contains 0.8*(20-x) quartz of the pure antifreeze.
Adding water does not change the amount of the antifreeze in the solution.
At the end, the volume of the pure antifreeze in the radiator after adding x quartz of water is 0.5*20 quartz.
So, we equate these two expressions for the pure antifreeze amount
0.8*(20-x) = 0.5*20 quartz. (1)
Simplify and find x
16 - 0.8x = 10,
16 - 10 = 0.8x,
6 = 0.8x
x = = = 7.5.
ANSWER. 7.5 quartz of the original 80% solution should be drained and replaced by pure water to get the 50% antifreeze solution.
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