Pipe D can fill a tank in $15$ hours, and pipe E can fill the same tank in $18$ hours. Pipe F can empty the tank in $25$ hours. Initially, pipes D and E are opened together for $5$ hours. After that, pipe E is closed, and pipe F is opened along with pipe D. Find the total time taken by pipes D and F together to fill the remaining part of the tank.

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**Solution:**

1. Calculate the rate at which each pipe fills or empties the tank:
   - Pipe D fills the tank in $15$ hours, so its rate is $\frac{1}{15}$ tank/hour.
   - Pipe E fills the tank in $18$ hours, so its rate is $\frac{1}{18}$ tank/hour.
   - Pipe F empties the tank in $25$ hours, so its rate is $-\frac{1}{25}$ tank/hour (negative because it empties).

2. When pipes D and E are opened together, their combined filling rate is:
   $$\frac{1}{15} + \frac{1}{18} = \frac{6 + 5}{90} = \frac{11}{90} 	ext{ tank/hour}$$

3. They work together for $5$ hours, so the amount of tank filled is:
   $$5 	imes \frac{11}{90} = \frac{55}{90} = \frac{11}{18}$$

4. Remaining part of the tank to be filled:
   $$1 - \frac{11}{18} = \frac{7}{18}$$

5. Now, pipe E is closed and pipe F is opened along with pipe D. Their combined rate is:
   $$\frac{1}{15} - \frac{1}{25} = \frac{5 - 3}{75} = \frac{2}{75} 	ext{ tank/hour}$$

6. Time taken by pipes D and F together to fill the remaining $\frac{7}{18}$ tank is:
   $$	ext{Time} = \frac{	ext{Work}}{	ext{Rate}} = \frac{\frac{7}{18}}{\frac{2}{75}} = \frac{7}{18} 	imes \frac{75}{2} = \frac{7 	imes 75}{18 	imes 2} = \frac{525}{36} = 14 \frac{7}{12} 	ext{ hours}$$

**Answer:** $14 \frac{7}{12}$ hours

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