Glass A contains 100 ml of water and glass B contains 100 ml of wine. A 10 ml spoonful of wine is taken from glass B and mixed thoroughly with...

After adding 10 mL of wine to the water, glass A contains 110 mL of liquid at a 10:1 ratio of water to wine, by volume. Since the solution is mixed thoroughly, we can assume that the distribution of wine in the water is constant.

So ten parts of a 10 mL sample is water and one part of it is wine. (10/11)*10mL water + (1/11)*10mL wine = 10 mL.

Glass A now contains (10/11)*100 mL water + (1/11)*100 mL wine.

Glass B now contains (10/11)*10 mL water + (1/11)*10mL wine + 90 mL wine

Glass A now contains 100/11  mL of wine and Glass B contains 100/11 mL of water. They are the same.