SOLUTION: A radiator can contain 6 gallons of liquid water and antifreeze. How many gallons of pure water would be needed to be mixed with 80% antifreeze to fill up the radiator with a 30%

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Question 888095: A radiator can contain 6 gallons of liquid water and antifreeze. How many gallons of pure water would be needed to be mixed with 80% antifreeze to fill up the radiator with a 30% antifreeze solution?
Found 3 solutions by josgarithmetic, lwsshak3, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!
u how many gallons water
v how many gallons of 80% antifreeze

;
solve for v.

Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website!
A radiator can contain 6 gallons of liquid water and antifreeze. How many gallons of pure water would be needed to be mixed with 80% antifreeze to fill up the radiator with a 30% antifreeze solution?
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let x=amt of pure water to mix
6-x=amt of 80% antifreeze to mix
..
80%(6-x)=30%*6
4.8-.8x=1.8
.8x=3
x=3.75
6-x=2.25
gallons of pure water would be needed to be mixed with 80% antifreeze=3.75

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!

A radiator can contain 6 gallons of liquid water and antifreeze. How many gallons of pure water would be needed to be mixed with 80% antifreeze to fill up the radiator with a 30% antifreeze solution?

Water to be mixed: , or gallons