SOLUTION: A radiator contains 10 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 10 gal

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Question 709907: A radiator contains 10 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 10 gal of a 50% antifreeze solution?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Try to start with the expression for what is in the radiator at start.

%2810%2A20%29%2F10=20
The righthand member is 20% antifreeze, and the left hand member is the quantity of pure antifreeze over 10 gallons of the liquid in the radiator.

In the numerator, you want to SUBTRACT g gallons of the 20% liquid and ADD g gallons of the 100% material.
10%2A20-g%2A20%2Bg%2A100.
The denominator will stay unchanged, 10, for 10 gallons. Also, you want this used as a ratio equalling 50, for 50% antifreeze.

%2810%2A20-g%2A20%2Bg%2A100%29%2F10=50
Solve for g.

%28200%2B%28100-20%29g%29=50%2A10
200%2B80g=500
80g=300
g=30/8=15/4
highlight%28g=3%263%2F4%29 gallons to remove and replace with 100% antifreeze.