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) (Show Source): You can put this solution on YOUR website!
Try to start with the expression for what is in the radiator at start.
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.
.
The denominator will stay unchanged, 10, for 10 gallons. Also, you want this used as a ratio equalling 50, for 50% antifreeze.
Solve for g.
g=30/8=15/4
gallons to remove and replace with 100% antifreeze.
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