A group of 12 men can do a piece of work in 14 days and other group of 12 women can do the same work in 21 days. They begin together but 3 days before the completion of work, man's group leaves off. The total number of days to complete the work is:

Answer: Option C

Let x be the required number of days
Given,
12 men and 12 women can complete a work separately in 14 days and 21 days respectively
Then,
12 men's 1 day work = $$\frac{1}{14}$$
And,
12 women's 1 day work = $$\frac{1}{21}$$
Then ,
12 women's 3 days work = $$\frac{3}{21}$$ = $$\frac{1}{7}$$

The remaining work = $$1-\frac{1}{7}$$  = $$\frac{6}{7}$$
Man's group leaves 3 days before the completion of work
That is, they were working together for x - 3 days
Thus, we have $$\frac{1}{7}$$ work left to be done in last 3 days by the women's group. This also means $$\frac{6}{7}$$ th of work has been done by both the groups (before men left)
Now, (12 men + 12 women)'s 1 day work = $$\frac{1}{14}+\frac{1}{21}$$   = $$\frac{5}{42}$$
i.e., $$\frac{5}{42}$$ work is done by 2 groups in 1 day.
So, $$\frac{6}{7}$$ of work is done by 2 groups together in $$\frac{42}{5}\times \frac{6}{7}$$   = $$\frac{36}{5}$$ days
Total time take to complete the work will be
= $$\frac{36}{5}$$ + 3 = $$\frac{51}{5}$$