SOLUTION: A radiator contains 10 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 55% antifreeze? (R

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Question 268025: A radiator contains 10 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 55% antifreeze? (Round your answer to two decimal places.)
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In words:
(antifreeze started out with - antifreeze removed + antifreeze added)/(total amout of fluid finally in radiator) = 55%
Let x = amount of fluid to be drained
given:
Radiator started with:
10 quarts of fluid
10%2A.25+=+2.5 quarts antifreeze
.25x = quarts antifreeze drained
x = quarts of pure antifreeze added
-----------------------------------------
%282.5+-+.25x+%2B+x%29%2F10+=+.55
(Note that radiator stated with 10 quarts and ended with 10 quarts)
%281.25x+%2B+2.5%29%2F10+=+.55
1.25x+%2B+2.5+=+5.5
1.25x+=+3
x+=+2.4
2.4 quarts of fluid must be drained and replaced with pure antifreeze
check answer:
%282.5+-+.25x+%2B+x%29%2F10+=+.55
%282.5+-+.25%2A2.4+%2B+2.4%29%2F10+=+.55
%282.5+-+.6+%2B+2.4%29%2F10+=+.55
5.5%2F10+=+.55
5.5+=+5.5
OK