A can complete 25\% of a work in 7 days, while B can complete 2/7 of the same work in 12 days. A and B start working together and continue for 12 days. After that, both are replaced by C, who finishes the remaining work in 3 days. In how many days can C complete the entire work alone?

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**Solution:**

1. Calculate the total time taken by A to complete the entire work:
   - Since A does 25\% of the work in 7 days, the full work will take:
     $$\frac{7}{0.25} = 28 \text{ days}$$

2. Calculate the total time taken by B to complete the entire work:
   - B completes $ \frac{2}{7} $ of the work in 12 days.
   - Therefore, total time for B is:
     $$\frac{12}{\frac{2}{7}} = 12 \times \frac{7}{2} = 42 \text{ days}$$

3. Calculate the combined work done by A and B in 1 day:
   $$\frac{1}{28} + \frac{1}{42} = \frac{3}{84} + \frac{2}{84} = \frac{5}{84} \text{ work/day}$$

4. Work done by A and B together in 12 days:
   $$12 \times \frac{5}{84} = \frac{60}{84} = \frac{5}{7} \text{ of the work}$$

5. Remaining work after 12 days:
   $$1 - \frac{5}{7} = \frac{2}{7}$$

6. C completes the remaining $ \frac{2}{7} $ work in 3 days, so C's 1 day work is:
   $$\frac{2/7}{3} = \frac{2}{21} \text{ work/day}$$

7. Therefore, C alone can complete the entire work in:
   $$\frac{1}{\frac{2}{21}} = \frac{21}{2} = 10.5 \text{ days}$$

**Answer:** 10.5 days

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