Three people D, E and F are hired for a work. If F works with 200% of his efficiency, then the time taken by him is equal to the time taken by D (who works with normal efficiency). E takes 24 days to complete the work while D and E together can complete the whole work in 8 days. In how many days will D, E and F together (working with their original efficiencies) complete the whole work? 

Let, time taken by F to do the work with 100% efficiency = y days So, time taken by F to do the work with 200% efficiency = y/2 days Therefore, time taken by D to do the work = y/2 days Work done by D and E in one day = 2/y + 1/24 = (48 + y)/24y So, (48 + y)/24y = 1/8 Or, 384 + 8y = 24y Or, 16y = 384 Or, y = 24 Therefore, F will take 24 days to do the work with 100% efficiency, and D will take 12 days. Work done by all three in one day = 1/12 + 1/24 + 1/24 = (2 + 1 + 1)/24 = 4/24 = 1/6 Required number of days = 1 ÷ (1/6) = 6 days