SOLUTION: Help please in this problem: Mixture Problem A radiator contains 25 liters of water and antifreeze solution, of which 60% is antifreeze. How much of this solution should be drai

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Help please in this problem: Mixture Problem A radiator contains 25 liters of water and antifreeze solution, of which 60% is antifreeze. How much of this solution should be drai      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 683803: Help please in this problem: Mixture Problem
A radiator contains 25 liters of water and antifreeze solution, of which 60% is antifreeze. How much of this solution should be drained and replaced with water for new solution to be 40% antifreeze?
Ans. is 8.33 liters, but I dont know the solutions..

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The liters of antifreeze in the radiator is
+.6%2A25+=+15+ liters
Let +x+ = the number of liters to be drained and replaced with water
+.6x+ = liters of antifreeze in the mixture that is drained
---------------
The key is that you end up with the same liters you started with
+%28+15+-+.6x+%29+%2F+25+=+.4+
+15+-+.6x+=+.4%2A25+
+15+-+.6x+=+10+
+.6x+=+15+-+10+
+.6x+=+5+
+x+=+8.333+ liters
------------------
check answer:
+%28+15+-+.6x+%29+%2F+25+=+.4+
+%28+15+-+.6%2A8.333+%29+%2F+25+=+.4+
+%28+15+-+5+%29+%2F+25+=+.4+
+10%2F25+=+.4+
+10+=+10+
OK