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 =
60
10
=6 units/day Similarly, B's efficiency =
60
12
=5 units/day and C's efficiency =
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=