MAT Exam 2021 Paper

Concept:Each operation removes $3.65$ L of the mixture and replaces it with water, leaving $95\%$ of the wine behind each time.Explanation:After one operation, the fraction of wine remaining is $1-\frac{3.65}{73}=0.95$.After $n$ operations, wine left in the drum is $73(0.95)^n$ L.The total volume remains $73$ L, so the concentration of wine after $n$ operations is $\frac{73(0.95)^n}{73}=(0.95)^n$.We need this concentration to be less than $85\%$, so $(0.95)^n < 0.85$.Testing values: $0.95^2=0.9025$, $0.95^3=0.857375$, and $0.95^4=0.814506$.For $n=3$, the concentration is $85.7375\%$, which is still greater than $85\%$.For $n=4$, the concentration is $81.4506\%$, which is less than $85\%$.Hence, the minimum number of operations required is $4$.Answer:$4$ operations (Option B).