Quantity I: Working alone, B can complete the work in 20 days. C is twice as efficient as B and A takes 2 days more than it takes C to complete the work. Working together, in how much time would they be able to complete 7 such works? Quantity II: Working alone, A, B and C can do a work in 24, 30 and 40 days respectively. How long will it take them to complete the work if only A and B work for the first 6 days and then C started the work?

Quantity I: B takes 20 days, so B's rate = 1/20. C is twice as efficient, so C's rate = 2/20 = 1/10 (C takes 10 days). A takes 2 days more than C, so A takes 12 days (rate = 1/12). Combined rate of A, B, C = 1/12 + 1/20 + 1/10 = 5/60 + 3/60 + 6/60 = 14/60 = 7/30. For 7 works: Time = 7 ÷ (7/30) = 30 days. Quantity II: A works for 6 days at rate 1/24, completing 6/24 = 1/4 of work. Remaining work = 3/4. With A, B, C working together at rate 1/24 + 1/30 + 1/40 = 5/120 + 4/120 + 3/120 = 12/120 = 1/10, time needed = (3/4) ÷ (1/10) = 7.5 days. Total time = 6 + 7.5 = 13.5 days. Since 30 > 13.5, Quantity I > Quantity II.