A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 4 days. How long B alone would take to do the whole work?

Let total work to be done = 100 units

Work done by A in 20 days = $$\frac{80}{100} \times 100 = 80$$ units

A's efficiency = $$\frac{80}{20} = 4$$ units/day

Remaining work = 100 - 80 = 20 units

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

Now, A and B complete remaining work in 4 days

=> $$(4 + x) \times 4 = 20$$

=> $$4 + x = \frac{20}{4} = 5$$

=> $$x = 5 - 4 = 1$$

$$\therefore$$ Time taken by B to complete the whole work alone = $$\frac{100}{1} = 100$$ days

=> Ans - (B)