A can contains a mixture of two liquids X and Y in the ratio 7:5. When 9 liters of mixture is taken out and replaced with liquid Y, the ratio of X to Y becomes 7:9. How many liters of liquid the can originally has?

Let's denote the total volume of the mixture in the can as $V$ liters.

Initially, the ratio of liquids X and Y is 7:5. Therefore, the volume of liquid X is $\frac{7}{12}V$ and the volume of liquid Y is $\frac{5}{12}V$.

When 9 liters of the mixture is taken out, the volumes of X and Y taken out will be in the same ratio of 7:5. So, the volume of liquid X taken out is $\frac{7}{12} \times 9 = 5.25$ liters, and the volume of liquid Y taken out is $\frac{5}{12} \times 9 = 3.75$ liters.

After removing 9 liters of the mixture, the remaining volumes of X and Y in the can are:

• Volume of X: $\frac{7}{12}V - 5.25$
• Volume of Y: $\frac{5}{12}V - 3.75$

Next, 9 liters of liquid Y is added to the can. So, the new volume of liquid Y becomes: $\left( \frac{5}{12}V - 3.75 \right) + 9 = \frac{5}{12}V + 5.25$

According to the problem, the new ratio of X to Y is 7:9. Therefore, we can set up the following equation: $\frac{\frac{7}{12}V - 5.25}{\frac{5}{12}V + 5.25} = \frac{7}{9}$

Cross-multiplying to solve for $V$: $9 \left( \frac{7}{12}V - 5.25 \right) = 7 \left( \frac{5}{12}V + 5.25 \right)$

Expanding both sides: $9 \times \frac{7}{12}V - 9 \times 5.25 = 7 \times \frac{5}{12}V + 7 \times 5.25$ $\frac{63}{12}V - 47.25 = \frac{35}{12}V + 36.75$

Combining like terms: $\frac{63}{12}V - \frac{35}{12}V = 36.75 + 47.25$ $\frac{28}{12}V = 84$ $\frac{7}{3}V = 84$ $V = 84 \times \frac{3}{7}$ $V = 36$

Therefore, the can originally has 36 liters of the mixture.