We'll say 10% gets poured first and that A & B have the same volume.
At the first pour A to B, B_Glass is (1/10A + 10/10B)
Inorder to obtain 10% we'll take 1/11A
as it's percent in comparison with the rest of B_Glass.
When the second exchange takes place
A_Glass = (9/10A) + 1/10 times (1/11A + 10/11B)
which is (9/10A + 1/110A + 10/110B)
therefore when they are both half again
A_Glass = ((9x11)/(10x11)A + 1/110A + 10/110B) = (100/110A + 10/110B)
B_Glass = (10/110A + 100/110B)
so we've distributed just as much A in B as B in A.
Another way, if you poured all of A_Glass into B_Glass
and then poured half of that back into the A_Glass there would be an equal amount of A and B in both Glasses.
And actually passed?