A and B can separately do a piece of work in 20 and 15 days respectively. They worked together for 6 days after which B was replaced by C. The work was finished in next 4 days. The number of days in which C alone could do the work is :

Work done by A in 1 day= = $$\frac{1}{20}$$
Work done by B in 1 day= = $$\frac{1}{15}$$
A and B worked together for 6 days.Work completed in 6 days =$$6(\frac{1}{A} + \frac{1}{B})$$
$$=6(\frac{1}{20} + \frac{1}{15}) = 6\frac{7}{60} = \frac{7}{10}$$
A and C worked together for next 4 days to complete the remaining $$\frac{3}{10}$$ th of the work.
$$4(\frac{1}{20} + \frac{1}{C}) = \frac{3}{10}$$
$$\frac{1}{5} + \frac{4}{C} = \frac{3}{10}$$
$$\frac{4}{C} = \frac{3}{10} - \frac{1}{5}$$
$$\frac{4}{C} = \frac{1}{10}$$
$$\frac{1}{C} = \frac{1}{40}$$
Hence C alone would take 40 days to complete the entire work.
Option D is the correct answer.