A does 50% of a work in 12 days. He then calls in B and they together finish the remaining work in 8 days. How long B alone would take to complete the whole work?

Let total work to be done = 96 units

Work done by A in 12 days = $$\frac{50}{100} \times 96 = 48$$ units

A's efficiency = $$\frac{48}{12} = 4$$ units/day

Remaining work = 96 - 48 = 48 units

Let B's efficiency = $$x$$ units/day

Now, A and B complete remaining work in 8 days

=> $$(4 + x) \times 8 = 48$$

=> $$4 + x = \frac{48}{8} = 6$$

=> $$x = 6 - 4 = 2$$

$$\therefore$$ Time taken by B to complete the whole work alone = $$\frac{96}{2} = 48$$ days

=> Ans - (D)