A vessel contains $250$ liters of a mixture of alcohol and water in the ratio $7:3$. Then, $45$ liters of another mixture containing water and alcohol in the ratio $5:4$ is added to the vessel. What is the difference between the amount of water and alcohol in the final mixture?

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**Solution:**

1. Calculate the amount of alcohol and water in the initial $250$ liters mixture:
   - Alcohol = $\frac{7}{10} 	imes 250 = 175$ liters
   - Water = $\frac{3}{10} 	imes 250 = 75$ liters

2. Calculate the amount of water and alcohol in the $45$ liters added mixture:
   - Water = $\frac{5}{9} 	imes 45 = 25$ liters
   - Alcohol = $\frac{4}{9} 	imes 45 = 20$ liters

3. Find the total amount of alcohol and water after mixing:
   - Total alcohol = $175 + 20 = 195$ liters
   - Total water = $75 + 25 = 100$ liters

4. Calculate the difference between water and alcohol in the final mixture:
   $$	ext{Difference} = 195 - 100 = 95 	ext{ liters}$$

**Answer:** $95$ liters

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