A and B, working separately can do a piece of work in 12 days and 20 days respectively. If they work alternately, beginning with A, then in how many days, can the work be completed?

15

The correct answer is Option (1) → 15

A’s one-day work = $\frac{1}{12}$​
B’s one-day work = $\frac{1}{20}$​

Work done in 2 days (A + B):

$\frac{1}{12}+\frac{1}{20}=\frac{5+3}{60}=\frac{8}{60}=\frac{2}{15}$

Work done in 14 days (7 such cycles):

$7 \times \frac{2}{15}=\frac{14}{15}$

Remaining work after 14 days:

$1-\frac{14}{15}=\frac{1}{15}$

On the 15th day, A works alone.

Time taken by A to complete $\frac{1}{15}$​ work:

$\frac{1/15}{1/12}=\frac{12}{15}=\frac{4}{5}\text{ day}$

So the work is completed during the 15th day.