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?
Answer & explanation
Correct answer: option 1
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.