A, B and C alone can complete a work in 24, 48 and 32 days respectively. All of them started working together but after 2 days from start A left the job and after 3 more days B also left the job. So for how many days did C work?

Solution

Let the total work = 96 units (LCM of 24, 48 and 32) Amount of work done by A alone in one day = 96/24 = 4 units Amount of work done by B alone in one day = 96/48 = 2 units Amount of work done by C alone in one day = 96/32 = 3 units Amount of work done by A, B and C together in 2 days = 2 × (4 + 2 + 3) = 18 units Amount of work done by B and C together in 3 days = 3 × (2 + 3) = 15 units Remaining work = 96 – 18 – 15 = 63 units So, the time taken by C alone to complete 63 units work = 63/3 = 21 days So, C worked for 21 + 2 + 3 = 26 days