A container has a mixture of milk and water in the ratio 5:3. If 4 liters of the mixture are replaced with 4 liters of milk, the new ratio becomes 7:3. What is the initial quantity of the mixture? Options 35 liters 25 liters 20 liters 30 liters

Let's denote the initial quantity of the mixture as $x$ liters.

1. The initial ratio of milk to water is 5:3. Therefore, the initial quantities of milk and water can be expressed as:

   • Milk: $\frac{5}{8}x$
   • Water: $\frac{3}{8}x$

2. When 4 liters of the mixture are removed, the quantities of milk and water removed are:

   • Milk removed: $4 \times \frac{5}{8} = 2.5$ liters
   • Water removed: $4 \times \frac{3}{8} = 1.5$ liters

3. The remaining quantities of milk and water in the container are:

   • Milk: $\frac{5}{8}x - 2.5$
   • Water: $\frac{3}{8}x - 1.5$

4. After removing 4 liters of the mixture, 4 liters of milk are added. The new quantities of milk and water are:

   • Milk: $\frac{5}{8}x - 2.5 + 4 = \frac{5}{8}x + 1.5$
   • Water: $\frac{3}{8}x - 1.5$

5. The new ratio of milk to water is 7:3. Therefore, we can set up the following equation: $\frac{\frac{5}{8}x + 1.5}{\frac{3}{8}x - 1.5} = \frac{7}{3}$

6. Cross-multiplying to solve for $x$: $3 \left( \frac{5}{8}x + 1.5 \right) = 7 \left( \frac{3}{8}x - 1.5 \right)$ $3 \left( \frac{5}{8}x + 1.5 \right) = 7 \left( \frac{3}{8}x - 1.5 \right)$ $\frac{15}{8}x + 4.5 = \frac{21}{8}x - 10.5$

7. Simplifying the equation: $4.5 + 10.5 = \frac{21}{8}x - \frac{15}{8}x$ $15 = \frac{6}{8}x$ $15 = \frac{3}{4}x$

8. Solving for $x$: $x = 15 \times \frac{4}{3}$ $x = 20$

Therefore, the initial quantity of the mixture is 20 liters.

The correct option is: 20 liters