SOLUTION: A radiator contains 20 liters of a 40% antifreeze solution. How many liters of the solution must be drained and replaced with pure antifreeze to have a 50% solution in the radiator
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Question 516111: A radiator contains 20 liters of a 40% antifreeze solution. How many liters of the solution must be drained and replaced with pure antifreeze to have a 50% solution in the radiator?(Answer is decimal liters with 2 digits to the right of the decimal point.)
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
the quantity of antifreeze required to make 20 l to 50% is the amount you need to drain
let that be x l
anifreeze 100 ---------------- x l
Solution 40 ------ 20 - x l
Mixture 50.00% ---------------- 20
100x +40(20 - x ) = 50 * 20
100x+ 800- 40x = 1000
100x - 40x = 1000 - 800
60x = 200
/ 60
x= 3.33 l anifreeze
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