How do you set the up this mixture problem?

Question

The radiator in a car is filled with a solution of 60% antifreeze and 40% water. I want it to be 50% antifreeze 50% water. If the radiator holds 3.6L, how much coolant should be drained and replaced with water get the desired percent.

Answer 1

Right now you have .6 x 3.6 L = 2.16 L antifreeze

and .4 x 3.6 = 1.44 L water

You want 1.8 L of each.

So you need to remove (2.16 - 1.8) / .6 L of solution

or .36/.6 = .6 L

3.6 - .6 = 3

3 x .6 = 1.8 L of antifreeze

3 x .4 = 1.2 L of water

add .6 L of water and it comes up to 1.8

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