SOLUTION: A car radiator needs a 40​% antifreeze solution. The radiator now holds 21 liters of a 10​% solution. How many liters of this should be drained and replaced with​
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Question 1002373: A car radiator needs a 40% antifreeze solution. The radiator now holds 21 liters of a 10% solution. How many liters of this should be drained and replaced with 100% antifreeze to get the desired strength?
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
21 liters of 0.10=2.1 liters of pure antifreeze.
Want 21 liters *0.4=8.4 liters of pure antifreeze. Let x be the amount that is new.
Then in the radiator, there will be (21-x) liters of 10% and x liters of 100%
(21-x)*(0.1)+1(x)=8.4
2.1-0.1x+x=8.4
0.9x=6.3l
x=7 liters.
7 liters of new
14 liters of old
Total is 7+(14)(0.1)=7+1.4=8.4 liters
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