A can complete a work in 24 days, B in 36 days and C in 48 days. All three start working together. After 4 days, A leaves the work. B and C continue the work for 6 more days. The remaining work is completed by B alone. In how many days did B alone complete the remaining work?

A's 1-day work = 1/24 B's 1-day work = 1/36 C's 1-day work = 1/48 (A + B + C)'s 1-day work = 1/24 + 1/36 + 1/48 = 6/144 + 4/144 + 3/144 = 13/144 Work done in 4 days = 4 × 13/144 = 52/144 = 13/36 Remaining work = 1 - 13/36 = 23/36 Now, B + C work for 6 days. (B + C)'s 1-day work = 1/36 + 1/48 = 4/144 + 3/144 = 7/144 Work done in 6 days = 6 × 7/144 = 42/144 = 7/24 Remaining work after that = 23/36 - 7/24 = 46/72 - 21/72 = 25/72 B alone does 1/36 work per day. Required time = (25/72) ÷ (1/36) = (25/72) × 36 = 25/2 = 12.5 days Hence, the correct answer is 12.5 days.