If a portion of a half water/half alcohol mix is replaced with 25% alc

Question on Mixtures can be easily solved using weighted averages concept discussed here:
https://gmatclub.com/forum/tough-ds-105651.html#p828579
I would also recommend that you go through the complete theory from some standard book if you are not comfortable.

This question involves replacement rather than simply mixing two solutions hence it has one extra step at the end which I will discuss later.

First of all consider, you have a solution with 50% alcohol. Part of it is removed and mixed with a 25% alcohol solution. In effect, this is similar to mixing two solutions, one of 25% alcohol and other of 50% alcohol.


So 25% alcohol solution volume : 50% alcohol solution volume is 20:5 which is 4:1.
Since out of 5 parts of 50% alcohol solution, 4 parts were removed to replace them with 4 parts 25% alcohol solution, the portion of 50% alcohol solution replaced was 4/5 = 80%.
The question asks: what percentage of original alcohol was replaced? Since the solution is homogenous, it you replace 80% of it, 80% of the original amount of alcohol in the solution will be replaced.

To understand this, lets say I have 100 liters of 50% alcohol solution. When I remove 80% of it, I remove 80 liters solution. In the solution I remove, I have 40 liters alcohol and 40 liters water. In the 20 liters solution that is remaining, 10 liters is alcohol and 10 liters is water. So amount of alcohol removed (and replaced) is 40/50 = 80%