A, B and C can complete a work in 10, 12 and 15 days respectively. A left the work 5 days before the work was completed and B left 2 days after A had left. Number of days required to complete the whole work are:

Let total work to be done = L.C.M. (10,12,15) = 60 units

=> A can complete the work in 10 days, => A's efficiency = $$\frac{60}{10}=6$$ units/day

Similarly, B's efficiency = $$\frac{60}{12}=5$$ units/day

and C's efficiency = $$\frac{60}{15}=4$$ units/day

Let the work is completed in $$t$$ days

Thus, A worked for $$(t-5)$$ days and B worked for $$(t-3)$$ days

=> $$6(t-5)+5(t-3)+4(t)=60$$

=> $$(6t-30)+(5t-15)+(4t)=60$$

=> $$15t=60+30+15=105$$

=> $$t=\frac{105}{15}=7$$ days

=> Ans - (D)