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

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> 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       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

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) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the number of quarts drained and replaced.
After draining x quarts, you have 20-x 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
0.70%2A%2820-x%29 quarts.
After you add x quarts of antifreeze, you end up with 20 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
0.50%2A20 quarts = 10 quarts.
Since you did not add any water, all that water is exactly the same amount you had after the draining step, or
0.70%2A%2820-x%29 quarts.
So 10=0.70%2A%2820-x%29
10=0.70%2A%2820-x%29 --> 10%2F0.70=0.70%2A%2820-x%29%2F0.70 --> 10%2F0.70=20-x --> 10%2F0.70%2Bx-10%2F0.70=20-x%2Bx-10%2F0.70 --> x=20-10%2F0.70 --> highlight%28x=5.7%29 quarts (rounding)
NOTE: In x=20-10%2F0.70 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 "=".