SOLUTION: A radiator contains 7 quarts of fluid, 25% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 45% antifreeze?
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Question 706832: A radiator contains 7 quarts of fluid, 25% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 45% antifreeze? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of fluid that needs to be drained and replaced with pure antifreeze
Now we know that the amount of pure antifreeze that exists before the mixture takes place(0.25(7-x)+x) has to equal the amount of pure antifreeze that exists after the mixture takes place(0.45*7), soooo:
0.25(7-x)+x=0.45*7 simplify
1.75-0.25x+x=3.15
0.75x=3.15-1.75
0.75x=1.4
x=1.87 qts-------------ans
CK
0.25(7-1.87)+1.87=3.15
1.28+1.87=3.15
3.15=3.15
Hope this helps----ptaylor
