A bartender stole champagne from a bottle that contained 50% of 
A bartender stole champagne from a bottle that contained 50% of spirit and he replaced what he had stolen with champagne having 20% spirit. The bottle then contained only 25% spirit. How much of the bottle did he steal?

My approach: Let the total quantity champagne containing  (50% of the spirit) be x.
$\therefore$  it will contain x/2  spirit and x/2 non-spirit.
Suppose he steals y% from it. Let the total quantity of champagne containing(20 % of spirit) be z.
$\therefore$ It will contain z/5 spirit.
Now I'm not able to form an equation please help.

Without (much) math.
Must steal enough champagne so that weighted average goes 5/6 of the way from 50 to 20 
(i.e. $[25-20] \times 5 = [50 - 25]$).
Therefore, he must steal 5/6 of the bottle.
I mention this approach not as a recommended approach for someone new to the problem, but rather as a way of developing intuition.
After enough experience with weighted average problems, they become solvable without (much) math.
