A started a work and left after working for 2 days. Then B finished the remaining work in 9 days. Had A left the work after working for 3 days. B would have finished the remaining work in 6 days. In how many days B alone can finish the work ?

Let total work = W. A's 2 days + B's 9 days = W. Also, A's 3 days + B's 6 days = W. Subtracting: (A's 1 day) - (B's 3 days) = 0, so A's 1 day = B's 3 days. Substituting in first equation: 3(B's 3 days) + 9(B days) = W, 18(B days) + 9(B days) = 27(B days) = W. Therefore B alone takes 27 days. Wait, let me reconsider: if A works 2 days and B works 9 days to complete, and A works 3 days and B works 6 days to complete, then the difference is: A works 1 extra day, B works 3 fewer days. So 1 day of A = 3 days of B. From first equation: 2×(3 days of B) + 9 days of B = 15 days of B = total work. Therefore B alone takes 15 days. Options A, B, C, and E are incorrect.