Bunuel wrote:

Official Solution:

If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?

A. \(3\%\)

B. \(20\%\)

C. \(66\%\)

D. \(75\%\)

E. \(80\%\)

Initial solution is "half water/half alcohol mix" means that it's 50% alcohol solution.

Let the portion replaced be \(x\) and the volume of initial solution be 1 unit.

Then the amount of alcohol after removal of that portion will be \(0.5(1-x)\) and the amount of alcohol added will be \(0.25x\), so total amount of alcohol will be \(0.5(1-x)+0.25x\). On the other hand, since finally 30% alcohol solution was obtained then the amount of alcohol in the end was \(0.3*1\).

So \(0.5(1-x)+0.25x=0.3\) \(\rightarrow\) \(x=0.8\), or 80%.

Answer: E

Hi Bunuel

I tried this question in a different way

Portion A Portion B

50% 25%

30%

5% 20%

Thus the portion removed is 1/5 which is 20 percent. Where did i go wrong??