Rani and Adya, working separately, can finish a task in 12 days and 16 days, respectively. They work in stretches of one day alternately. If Rani starts working first, then the task will be completed in how many days?

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**Solution:**

1. Rani can complete the task in 12 days, so her one day work = $\frac{1}{12}$ of the task.

2. Adya can complete the task in 16 days, so her one day work = $\frac{1}{16}$ of the task.

3. They work alternately one day each, starting with Rani.

4. Work done in 2 days (Rani + Adya) = $\frac{1}{12} + \frac{1}{16} = \frac{4}{48} + \frac{3}{48} = \frac{7}{48}$ of the task.

5. In 12 days, there are 6 such 2-day cycles, so work done in 12 days = $6 	imes \frac{7}{48} = \frac{42}{48} = \frac{7}{8}$ of the task.

6. Remaining work after 12 days = $1 - \frac{7}{8} = \frac{1}{8}$ of the task.

7. The next day is Rani's turn, who does $\frac{1}{12}$ of the work per day.

8. Time taken by Rani to finish remaining $\frac{1}{8}$ work = $\frac{\frac{1}{8}}{\frac{1}{12}} = \frac{1}{8} 	imes 12 = \frac{12}{8} = 1.5$ days.

But since the work is done in stretches of one day alternately, Rani works only one day at a time.

Actually, after 12 days (6 cycles), 7/8 work is done. Next day (13th day) Rani works and completes $\frac{1}{12}$ work.

Remaining work after 13 days = $\frac{1}{8} - \frac{1}{12} = \frac{3}{24} - \frac{2}{24} = \frac{1}{24}$.

Next day (14th day) Adya works and completes $\frac{1}{16}$ work, which is more than remaining $\frac{1}{24}$, so she finishes the task during her day.

Time taken by Adya to complete remaining $\frac{1}{24}$ work:

$$	ext{Time} = \frac{	ext{Work}}{	ext{Rate}} = \frac{\frac{1}{24}}{\frac{1}{16}} = \frac{1}{24} 	imes 16 = \frac{2}{3} 	ext{ day}$$

Total time = 13 days (full days) + $\frac{2}{3}$ day = $13 \frac{2}{3}$ days.

**Answer:** $13 \frac{2}{3}$ days

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