A and B can do a piece of work in 30 days and 18 days respectively. A started the work alone and then after 6 days B joined him till the completion of the work. In how many days has the whole work completed?

Let the total work be W

Number of days required for A to complete the work = 30 days

$$\Rightarrow$$  Work done by A in 1 day = $$\frac{W}{30}$$

Number of days required for B to complete the work = 18 days

$$\Rightarrow$$  Work done by B in 1 day = $$\frac{W}{18}$$

Work done by A and B together in 1 day = $$\frac{W}{30}+\frac{W}{18}$$ = $$\frac{3W+5W}{90}$$ = $$\frac{4W}{45}$$

Work done by A alone in 6 days = $$\frac{W}{30}\times6$$ = $$\frac{W}{5}$$

Remaining work = $$W-\frac{W}{5}$$ = $$\frac{4W}{5}$$

Number of days required for both A and B to complete remaining work = $$\frac{\text{Remaining work}}{\text{Work in 1 day}}$$ = $$\frac{\frac{4W}{5}}{\frac{4W}{45}}$$ = 9 days

$$\therefore\ $$Number of days required to complete the whole work = 6 + 9 = 15 days

Hence, the correct answer is Option B