# Mix-problem with percentage

A can is containing coffee and another can is containing exactly the same amount of milk. We take a spoon of coffee and mix it in the milk can. Then we take a spoon from the mix we obtained and mix it in the coffee. Now we have obtain two different mixes, one with some coffee in the milk, and one with some milk in the coffee. Is it more milk in the coffee than it's coffee in the milk or is it the same?

I thought like this: We name the amount in the cans x. Then it lasts x-1 in the coffee-can and the milk-can contains x+1 after the first mix. And then we take $\frac{1}{x+1}$ of this mixture in the coffee-can.

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Let each can contain $V$. You start with $V$ coffee in one can and $V$ milk in the other. After some spooning (it doesn't matter how much or whether you mix), the coffee can has $c$ coffee and $m=V-c$ milk. The milk can has $V-m=c$ milk and $V-c$ coffee. So there is the same amount of coffee in the milk as milk in the coffee.
So one has C-T coffee and the other one has C milk. Then you take away a teaspoon from the second one, However this teaspoon only has $t\frac{C}{C+T}$ milk in it. Since \$\frac{C}{C+T} is less than one then that means that the cup of milk lost less than one teaspoon of milk. And since the two cups contain the same amount of liquid and the second one has more milk that means the cup of milk is purer. (note this only works when it is well stirred.)