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December 22, 2014

December 22, 2014

Posted by **coried** on Wednesday, August 5, 2009 at 3:27pm.

Radiator in a certain make needs to contain 50 liters of 40% antifreeze. The radiator now contains 50 liters of 20% antifreeze. How many liters of this solution must be drained and replaced with 100% antifreeze to get the desired strength?

- Algebra -
**drwls**, Wednesday, August 5, 2009 at 4:16pmLet X be the number of liters that must be drained and replaced with 100% antifreeze.

[0.2(50-X)+X]/50 = 0.4

Solve that for X.

- Algebra -
**bobpursley**, Wednesday, August 5, 2009 at 4:17pmYou want to double the concentration?

If you remove X liters, you remove .2X antifreeze, and .8X water.

You have left in the radiator after draining X, 10-.2X liters of pure antifreeze. You need a total of 20 liters to get 40percent antifreeze.

So you add 10+.2X of antifreeze.

You drain out 10-2X of solution.

X=10+.2X

.8X=10

x=12.5 liters

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