‘A’ can complete a work in 20 days while B is 25\% more efficient than ‘A’. B worked for 6 days and left, remaining work is completed by ‘C’ in 15 days. Find in how many days ‘C’ can complete the whole work alone?

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**Solution:**

1. Given, A can complete the work in 20 days.

2. B is 25\% more efficient than A, so B's efficiency = 1.25 × A's efficiency.

3. Since A completes work in 20 days, A's work rate = $ \frac{1}{20} $ work/day.

4. Therefore, B's work rate = 1.25 × $ \frac{1}{20} $ = $ \frac{1.25}{20} = \frac{5}{80} = \frac{1}{16} $ work/day.

5. B worked for 6 days, so work done by B = 6 × $ \frac{1}{16} $ = $ \frac{6}{16} = \frac{3}{8} $.

6. Remaining work = 1 - $ \frac{3}{8} $ = $ \frac{5}{8} $.

7. C completed the remaining work in 15 days, so C's work rate = $ \frac{5/8}{15} = \frac{5}{8 \times 15} = \frac{5}{120} = \frac{1}{24} $ work/day.

8. Hence, C alone can complete the whole work in 24 days.

**Answer:** 24 days

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