A and B together can do a piece of work in 30 days, B and C together can do it in 20 days, A starts the work and works on it for 5 days, B takes up and work for 15 days. Finally C finishes the work in 18 days. The number of days in which C alone can do the work where doing it separately is = ?

Answer: Option D

L.C.M. of Total Work =60
One day work of A + B = $$\frac{60}{30}$$ = 2 unit/day
One day work of B + C = $$\frac{60}{20}$$ = 3 unit/day
$$\begin{array}{rl}& \text{According to the question,}\\ & \Rightarrow \left(\text{A}+\text{B}\right).....\text{30 days}\\ & \Rightarrow \left(\text{B}+\text{C}\text{.}\right).....\text{20 days}\\ & \Rightarrow \text{A}.....\text{5}\\ & \Rightarrow \text{B}.....\text{5}+\text{10}\\ & \Rightarrow \text{C}.....\text{10}+\text{8 days}\end{array}$$
Work done by (A +B) in 5 days
= 2 × 5 = 10 units
⇒ Work done by (B + C) in 10 days
= 10 × 3 = 30 units
⇒ Total work (finished till now).....40 units
⇒ Remaining work
= 60 - 40
= 20 units
⇒ Here we find that C does remaining 20 units in 8 days
⇒ C's efficiency = $$\frac{\text{Work}}{\text{Day}}$$
$$=\frac{\text{20 Work}}{\text{8 Days}}=\frac{5}{2}$$
⇒ Therefore, time taken by C alone to complete the work
$$=\frac{60}{5}\times 2=24\text{ days}$$