A, B and C alone can complete a work in 30, 60 and 45 days respectively. All of them started working together but after 3 days from start A left the job and after 7 more days B also left the job. So for how many days did C work?

Solution

Let the total work = 180 units (LCM of 30, 60 and 45) Amount of work done by A alone in one day = 180/30 = 6 units Amount of work done by B alone in one day = 180/60 = 3 units Amount of work done by C alone in one day = 180/45 = 4 units Amount of work done by A, B and C together in 3 days = 3 × (6 + 3 + 4) = 39 units Amount of work done by B and C together in 7 days = 7 × (3 + 4) = 49 units Remaining work = 180 – 39 – 49 = 92 units So, the time taken by C alone to complete 92 units work = 92/4 = 23 days So, C worked for 23 + 3 + 7 = 33 days