SOLUTION: Please help me with this mixture problem. I have no idea how to do it. THanks A car radiator contains 10 liters of a 30% antifreeze solution. how many liters will have to be repla

Algebra ->  Algebra  -> Polynomials-and-rational-expressions -> SOLUTION: Please help me with this mixture problem. I have no idea how to do it. THanks A car radiator contains 10 liters of a 30% antifreeze solution. how many liters will have to be repla      Log On

Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 7860: Please help me with this mixture problem. I have no idea how to do it. THanks
A car radiator contains 10 liters of a 30% antifreeze solution. how many liters will have to be replaced with pure antifreeze if the resulting solution is to be 50% antifreeze?
Answer by Earlsdon(6287) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = the number of liters of antifreeze that must be replaced. Change percentages to decimals.
Start with 10 liters of 30% antifreeze. Write this as: 0.3(10)
You want to remove x liters of 30% antifreeze. Write this as -0.3(x)
Then you want add x liters of 100% antifreeze. Write this as +1(x) or just +x.
Now all of this must add up to 10 liters of 50% antifreeze. Write this as 0.5(10).
Putting it all together, we have:
0.3%2810%29+-+0.3x+%2B+x+=+0.5%2810%29 Simplify and solve for x.
3+%2B+0.7x+=+5
0.7x+=+2
x+=+2.86+
You must remove 2.86 liters of 30% antifreeze then add 2.86 liters of 100% antifreeze.