SOLUTION: The radiator of a jeep has a capacity of 4 gallons. It is filled with an anti-freeze solution of water and glycol which analyzes 10% glycol. What volume of the mixture must be draw

Question 551396: The radiator of a jeep has a capacity of 4 gallons. It is filled with an anti-freeze solution of water and glycol which analyzes 10% glycol. What volume of the mixture must be drawn off and replaced with glycol to obtain a 25% glycol solution? All percentages are by volume. Please thank you~ <3
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Think in terms of the amount of glycol in this situation.
Initially you have 4 gallons of solution that is 10% glycol.
The amount of glycol is then 4(0.1) = 0.4 gallons.
You want to remove some of this solution, let's call it x gallons. So we now have:
4(0.1)-x(0.1) of glycol.
Then you want to replace the removed amount of the 10% solution with the same amount of pure (100%) glycol or x(1), so we have:
4(0.1)-x(0.1)+x and this is to equal 4 gallons of 25% glycol solution which is 4(0.25) glycol.
So here's the equation to solve for x.
4(0.1)-x(0.1)+x = 4(0.25) Simplify.
0.4+0.9x = 1
0.9x = 0.6
x = 0.67 gallons.
You must draw off of a gallon of the 10% solution and replace it with of a gallon of pure glycol.
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