M can complete a work in 14 days less than the time taken by L. If both M and L together can complete the same work in 24 days, then in how many days L alone can complete the same work?

Let the number of days taken by L to complete the work be L days
Then, 1 day work of L = $$\dfrac{1}{L}$$

Number of days taken by M to complete the work = L-14 days
Then, 1 day work of M = $$\dfrac{1}{L-14}$$

Given, $$\dfrac{1}{L} + \dfrac{1}{L-14} = \dfrac{1}{24}$$

=> $$\dfrac{2L-14}{L^2-14L} = \dfrac{1}{24}$$

=> $$L^2 - 62L + 336 = 0$$
=> $$L^2 - 6L-56L+336 = 0$$
=> $$L(L-6)-56(L-6) = 0$$
=> $$(L-56)(L-6) = 0$$
=> L = 56 or L = 6
Since, M can do the work 14 days less than L, L cannot be 6.
Hence, L can do the work in 56 days.