R and S can comple1e a work in 15 days and 60 days respectively. They work on alternate days one at a time and R works on the first day. In how many days the whole work will be completed?

Let the total work be 1 unit.

Daily efficiencies:
R alone completes $$\frac{1}{15}$$ of the work per day, while S alone completes $$\frac{1}{60}$$ of the work per day.

They work on alternate days, with R working on Day 1, S on Day 2, R on Day 3, and so on. Therefore every pair of consecutive days constitutes one complete cycle:

Work finished in one 2-day cycle:
$$\frac{1}{15}+\frac{1}{60}=\frac{4}{60}+\frac{1}{60}=\frac{5}{60}=\frac{1}{12}$$

Thus, after $$n$$ such cycles (that is, after $$2n$$ days) the fraction of work completed is $$\frac{n}{12}$$.

To finish the work we need $$\frac{n}{12}\ge 1$$ ⟹ $$n\ge 12$$. Hence 12 full cycles are required.

Checking just before the last cycle:
• After 11 cycles (22 days) work done = $$\frac{11}{12}$$.
• Remaining work = $$1-\frac{11}{12}=\frac{1}{12}$$.

Day 23 (R’s turn): R can finish $$\frac{1}{15}$$. This is smaller than the remaining $$\frac{1}{12}$$, so some work is still left.
Work completed after Day 23: $$\frac{11}{12}+\frac{1}{15} =\frac{55}{60}+\frac{4}{60} =\frac{59}{60}$$.
Remaining work = $$\frac{1}{60}$$.

Day 24 (S’s turn): S exactly completes $$\frac{1}{60}$$, finishing the job.

Total time taken = 24 days.

Option C which is: 24 days