A, B and C can complete a work in 10, 12 and 15 days respectively. All three of them starts together but after 2 days A leaves the job and B left the job 3 days before the work was completed. C completed the remaining work alone. In how many days was the total work completed?

A, B and C can complete a work in 10, 12 and 15 days respectively.

Let total work to be done = L.C.M.(10,12,15) = 180 units

=> A's efficiency = $$\frac{180}{10}=18$$ units/day

Similarly B's efficiency = $$\frac{180}{12}=15$$ units/day

and C's efficiency = $$\frac{180}{15}=12$$ units/day

All of them works for 2 days, => Work done in 2 days = $$(18+15+12)\times2=90$$ units

C alone work for the last three days, => Work done by C alone = $$12\times3=36$$ units

Work left = $$180-90-36=54$$ units

Now, time taken by B and C together to complete 54 units of work = $$\frac{54}{(15+12)}=2$$ days

$$\therefore$$ Total time taken = $$3+2+2=7$$ days

=> Ans - (C)