SOLUTION: A radiator contains 5 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 50% antifreeze?

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Question 847174: A radiator contains 5 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 50% antifreeze?
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Drain v quarts of the 25% and replace with v quarts of the pure, 100%.

Amount of antifreeze will be: 5%2A25-v%2A25%2Bv%2A100, which is 100 times larger because of using the percents unit. Simplify partway into 25%2A5%2B%28100-25%29v.

The equation for wanting 50% antifreeze result is:
highlight%28%2825%2A5%2B%28100-25%29v%29%2F5=50%29.
Solve for v.

A next step could be the combined step, 125%2B75v=250....
75v=125
v=125%2F75=%285%2A5%2A5%29%2F%283%2A5%2A5%29
highlight%28v=5%2F3=1%262%2F3%29 quarts