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

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Question 667348: 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 lwsshak3(11628) About Me  (Show Source):
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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?
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let x= amt of 30% antifreeze to be drained and replaced with pure antifreeze
16-x= amt of 30% antifreeze remaining in radiator
30%(16-x)+100%x=50%*16
4.8-.3x+x=8
.7x=3.2
x=3.2/.7
x≈4.57
amt of 30% antifreeze to be drained and replaced with pure antifreeze=4.57 liters