[Head416]

I think what he's getting at is that even without a uniform mixture, the amount of water/ink will be the same, even if you can't say what that percentage is.

Say you take 1 ounce of ink and put it in the water, but do not mix it. Then you take 1 ounce out of this mixture, but you have no way of knowing what percentage of that ounce is ink. So, if this ounce has x ounces of ink (some fraction of 1 ounce) then the eyedropper has (1-x) ounces of water. When you add it to the ink, you now have (9+x) ounces of ink and (1-x) ounces of water. (9+x)+(1-x)=10 ounces

So in the water you had 10 ounces, added 1 ounce of ink, then removed x ounces of ink and (1-x) ounces of water. So the amount of water is [10-(1-x)]=(9+x) and the amount of ink is (1-x).