MAT Exam 2020 Paper

Concept:When a fixed volume is removed and replaced with another liquid, the total mixture remains at $50$ L. The water left after each process is best found using the formula: Final amount $= \text{Initial} \times \left(1 - \frac{\text{volume removed}}{\text{total volume}}\right)^n$, where $n$ is the number of times the process is repeated.Explanation:Start with water $= 50$ L.Each time, $5$ L is removed out of $50$ L, so the fraction removed is $\frac{5}{50} = \frac{1}{10}$.Since the process is repeated twice, water remaining in the cistern is:$50 \times \left(1 - \frac{1}{10}\right)^2 = 50 \times \left(\frac{9}{10}\right)^2 = 50 \times \frac{81}{100} = 40.5$ L.The total volume of the mixture never changes and remains $50$ L.Therefore, wine present in the final mixture is:$50 - 40.5 = 9.5$ L.Required proportion of wine to water is:$9.5 : 40.5$Multiplying both terms by $10$ gives $95 : 405$.Dividing both terms by $5$ gives the simplest ratio $19 : 81$.Answer:Option C: $19:81$.