A, B and C, alone can do a piece of work in 9, 12 and 18 days respectively. They all started the work together, but A left after 3 days. In how many days, was the remaining work completed ?

Let the Total Work = $$W$$

Work done by A in 1 day = $$\frac{W}{9}$$

Work done by B in 1 day = $$\frac{W}{12}$$

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

Work done by A,B,C in 3 days = $$3\times\frac{W}{9}+3\times\frac{W}{12}+3\times\frac{W}{18}=\frac{W}{3}+\frac{W}{4}+\frac{W}{6}=\frac{9W}{12}=\frac{3W}{4}\ $$

Remaining Work after A left(after 3 days) = $$W-\frac{3W}{4}=\frac{W}{4}$$

Work done by B and C in 1 day = $$\frac{W}{12}+\frac{W}{18}=\frac{5W}{36}$$

$$\therefore\ $$Number of days required to complete $$\frac{W}{4}$$ work by B and C = $$\frac{\frac{W}{4}}{\frac{5W}{36}}=\frac{W}{4}\times\frac{36}{5W}=\frac{9}{5}$$

Hence, the correct answer is Option B