If A, B and C can complete a piece of work in 6 days. If A can work twice faster than B and thrice faster then C, than the number of days C alone can complete the work is ?

Answer: Option B

A + B + C = 6 days [6 days total work]
$$\begin{array}{rl}& \text{According to the question,}\\ & \text{Ratio of their efficiencies,}\\ & \text{A}:\text{B}:\text{C}\\ & 6:3:2\\ & \text{Total efficiencies}\\ & \left(6+3+2\right)\text{units}=11\text{ units}\\ & \text{Total work}=11\times 6=66\text{ units}\end{array}$$
Therefore, time taken by C to complete the work
$$\frac{\text{Total work}}{\text{Efficiencies}}=\frac{66}{2}=33\text{ days}$$