SOLUTION: I am in college algebra. My problem is: A mechanic finds the car with a 20 quart radiator has a mixture containing 30% antifreeze in it. How much of this mixture would he have

Question 640207: I am in college algebra. My problem is:
A mechanic finds the car with a 20 quart radiator has a mixture containing 30% antifreeze in it. How much of this mixture would he have to drain out and replace with pure antifreeze to get a 50% antifreeze mixture?

Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
Let be the number of quarts drained and replaced.
After draining quarts, you have quarts of 30% solution in the radiator. That solution is 30% antifreeze and 70% water (100%-30%=70%).
The amount of water in the radiator at that point is
quarts.
After you add quarts of antifreeze, you end up with quarts of solution, that is supposed to be 50% antifreeze and 50% water.
The amount of water in the radiator at that point is supposed to be 50% of 20 quarts, or
quarts = quarts.
Since you did not add any water, all that water is exactly the same amount you had after the draining step, or
quarts.
So
--> --> --> --> --> quarts (rounding)
NOTE: In or x=20-10/0.70, according to order of operations rules (conventions) the division is done first. Most calculators know that, and will do it that way if you enter "20-10/0.7" before pressing "=".
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