A, B and C alone can complete a work in 12, 24 and 20

Solution

Let the total work = 120 units (LCM of 12, 24 and 20) Amount of work done by A alone in one day = 120/12 = 10 units Amount of work done by B alone in one day = 120/24 = 5 units Amount of work done by C alone in one day = 120/20 = 6 units Amount of work done by A, B and C together in 2 days = 2 × (10 + 5 + 6) = 42 units Amount of work done by B and C together in 3 days = 3 × (5 + 6) = 33 units Remaining work = 120 – 42 – 33 = 45 units So, the time taken by C alone to complete 120 units work = 45/6 = 7.5 days So, C worked for 7.5 + 2 + 3 = 12.5 days