A container contains 60 litres of a mixture of milk and water in the ratio 7:3. How many litres of the mixture should be removed and replaced with pure water so that the ratio of milk to water in the resulting mixture becomes 7:5?

Solution

Solution: Initially, milk \(=\frac{7}{10}\times60=42\) L and water \(=\frac{3}{10}\times60=18\) L. Let x litres of mixture be removed. Milk remaining \(=42-\frac{7x}{10}\). After removing the mixture and adding x litres of pure water, water becomes \(18-\frac{3x}{10}+x=18+\frac{7x}{10}\). Using the required ratio, \(\frac{42-\frac{7x}{10}}{18+\frac{7x}{10}}=\frac75\). Solving gives \(x=10\) litres. Trick: In replacement questions, the removed mixture contains milk and water in the original ratio; the replacement adds only water.