A can do a piece of work in 12 days and B in 20 days. If they together work on it for 5 days, and remaining work is completed by C in 3 days, then in how many days can C do the same work alone ?

Solution(By Examveda Team)

L.C.M. of Total Work =60 unit
One day work of A = $$\frac{{60}}{{12}}$$ = 5 unit/day
One day work of B = $$\frac{{60}}{{20}}$$ = 3 unit/day
$$\eqalign{ & {\text{5}}\left( {{\text{A}} + {\text{B}}} \right) + {\text{3C}} = {\text{60 units}} \cr & {\text{5}} \times {\text{8}} + {\text{3C}} = 60 \cr & 3{\text{C}} = 20 \cr & {\text{C}} = \frac{{20}}{3}{\text{ units/day}} \cr & {\text{Time taken by C}} \cr & = \frac{{60}}{{\frac{{20}}{3}}} \cr & = \frac{{60 \times 3}}{{20}} \cr & = 9{\text{ days}} \cr} $$