SOLUTION: An automobile radiator contains 16 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that t

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: An automobile radiator contains 16 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that t      Log On


   



Question 842207: An automobile radiator contains 16 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that the mixture will be 50% antifreeze?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+.3%2A16+=+4.8+ liters of
antifreeze in radiator initially
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Let +x+ = liters of mixture to be
removed and replaced with pure antifreeze
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+.3x+ = liters of pure antifreeze removed
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In words:
( liters of antifreeze you end up with ) / ( liters of mixture you end up with ) = 50%
+%28+4.8+-+.3x+%2B+x+%29+%2F+16+=+.5+
+4.8+%2B+.7x+=+.5%2A16+
+.7x+=+8+-+4.8+
+.7x+=+3.2+
+x+=+4.5714+ liters should be drained
and replaced with pure antifreeze
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check answer:
+.3%2A4.5714+=+1.3714+ liters of antifreeze removed
There is now +4.8+-+1.3714+=+3.4286+ liters
of antifreeze in radiator
+x+=+4.5714+ liters of antifreeze is put back in
There is now +3.4286+%2B+4.5714+=+8+ liters of
antifreeze in radiator and total of +16+ liters
+8%2F16+=+.5+ = 50%
OK