Let the total work be the least common multiple of 30, 24, and 20, which is 120 units. D's efficiency is 120 divided by 30, which is 4 units per day; E's efficiency is 120 divided by 24, which is 5 units per day; and F's efficiency is 120 divided by 20, which is 6 units per day. D and F work together for y days, completing (4 + 6)y = 10y units of work. E works alone for 4y days, completing 5 x 4y = 20y units of work. The total work is 10y + 20y = 30y units, which must equal the 120 units of total work; solving gives y = 4. The value of 2y is 2 x 4 = 8. The result is 8.