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 ?

Solution(By Examveda Team)

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