Question

A and B can do a work in 20 days. When A works at 60% capacity, B has to work at 150% capacity to finish the work. Find in how many days the faster one will finish the work alone.

Answer 1

Let x be the number of days the faster one (let's assume A is faster) will finish the work alone.

When A works at 60% capacity and B works at 150% capacity, their combined work rate is $\frac{3}{5}+\frac{3}{2}$​ of the normal capacity.

The equation $(\frac{3}{5}+\frac{3}{2}).\frac{1}{x}=\frac{1}{20}$​ is derived from the fact that A and B together can finish the work in 20 days.

Solving the equation, we find x=36.

Therefore, the faster one (A) will finish the work alone in 36 days.