A and B together can do a piece of work in 12 days, which B and C tog

Let us look at a method of solving this question without making equations using variables

A and B together can do a work in 12 days
B and C together can do it in 16 days
Let us assume the work to be a multiple of 12 and 16 = 48 units (to make our calculations easier)

Thus, work done by A and B together in 1 day = 48/12 = 4 units ... (i)
Work done by B and C together in 1 day = 48/16 = 3 units ... (ii)

We know that the work was completed in the following fashion:
A for 5 days, B for 7 days and C for 13 days
We can rearrange and regroup the work without changing the work done. We rearrange in such a way that we can group A and B together and then B and C together:

Total work
= Work done by A (5 days) + B (7 days) + C (13 days)
= Work done by A (5 days) + B (5 days) + B (2 days) + C (13 days)
= Work done by [A (5 days) + B (5 days)] + [B (2 days) + C (2 days)] + C (11 days)
= Work done by A and B together (5 days) + B and C together (2 days) + C (11 days)

Using (i) and (ii) and equating the total work, we have:
48 = (4 x 5) + (3 x 2) + C (11 days)
=> C (11 days) = 48 - 20 - 6 = 22
=> Work done by C in 1 day = 22/11 = 2

=> Time taken by C to complete the work alone = 48/2 = 24 days

Answer B