The container X contains 140 liters pure milk. $\frac{250}{7}$% of the milk is taken out from the container and 30 liters of water is added in it and then another 40% of the mixture is taken out. If another container Y has 80 liters of mixture in which 62.5% is milk and remaining soda, then find the final quantity of water in container X is what percent of quantity of soda in container Y ?

### Concept & Mixture Ratios This problem requires tracking the absolute quantities of specific components (milk and water) through multiple removal and addition steps. When a percentage of a mixture is removed, that exact percentage is removed from *each* component proportionately. ### Step-by-Step Solution **Step 1: Track the components in Container X** * Initial pure milk in X = $140$ liters. * First removal: $\frac{250}{7}\%$ of the milk is taken out. Fraction equivalent: $\frac{250}{7 \times 100} = \frac{250}{700} = \frac{5}{14}$. Amount removed = $\frac{5}{14} \times 140 = 50$ liters. Remaining milk = $140 - 50 = 90$ liters. * Addition: $30$ liters of water are added. New mixture volume = $90 \text{ (milk)} + 30 \text{ (water)} = 120$ liters. * Second removal: $40\%$ of the mixture is taken out. If 40% is taken out, $60\%$ remains. Since the mixture is uniform, exactly 60% of the water remains. Final quantity of water in X = $60\% \text{ of } 30 = \frac{60}{100} \times 30 = 18$ liters. **Step 2: Track the components in Container Y** * Total mixture in Y = $80$ liters. * Milk = $62.5\%$. Fraction equivalent of $62.5\% = \frac{5}{8}$. * Remaining part is soda, which is $100\% - 62.5\% = 37.5\%$. Fraction equivalent of $37.5\% = \frac{3}{8}$. * Quantity of soda in Y = $\frac{3}{8} \times 80 = 3 \times 10 = 30$ liters. **Step 3: Calculate the required percentage** * We need to find: (Final water in X) is what percent of (Soda in Y)? * Percentage = $(\frac{18}{30}) \times 100\%$ * Percentage = $(\frac{3}{5}) \times 100\% = 60\%$. ### Exam Strategy & Shortcut Memorize standard fraction-to-percentage conversions to speed up calculations: $62.5\% = \frac{5}{8}$ $37.5\% = \frac{3}{8}$ $\frac{250}{7}\% = \frac{5}{14}$ Using fractions instead of decimals for percentages like $\frac{250}{7}$ immediately simplifies $140 \times \frac{5}{14}$ into $50$ mentally. ### Common Pitfall A common mistake is trying to track the remaining milk during the second removal phase ($40\%$ removed). The question only asks for the final quantity of *water*, so calculating the final quantity of milk wastes valuable exam time. Focus only on the required variable. ### Final Answer Therefore, the correct answer is **60%**.