A can complete a job in twice the time that C and D take to complete the job, working together, and in 25% more time than that B and C take to complete the job, working together. If B is half as efficient as D and can complete the job in 40 days, working alone, then, in how many days can all three of them, working together, complete the job?

Let the work be 80 units, so D's efficiency is 2 units/day and B's efficiency is 1 unit/day. A takes 25% more time than B and C together, so A's efficiency is 4/5 of the combined efficiency of B and C, giving 4/5 x (1 + c) = a. Since A takes twice the time of C and D together, A's efficiency is half of their combined efficiency, giving 1/2 x (c + 2) = a. Equating the two expressions for A gives 4/5 x (1 + c) = 1/2 x (c + 2), which solves to c = 2 units/day. Substituting C's efficiency gives a = 2 units/day. The combined efficiency of A, B, and C is 2 + 1 + 2 = 5 units/day, so they complete the 80 units of work in 80 divided by 5, which is 8 days. The result is 8.

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