A and B working separately can do a piece of work in 9 and 15 days respectively. If they work for a day alternately, with A beginning, then the work will be complete in :
[A]9 days
[B]10 days
[C]11 days
[D]12 days

11 days
A’s 1 day’s work $= \frac{1}{9}$
B’s 1 day’s work $= \frac{1}{15}$
Work done in first 2 days = A’s 1 day’s work + B’s 1 day’s work
$= \frac{1}{9}+\frac{1}{15} = \frac{5+3}{45} = \frac{8}{45}$
∴ Work done in first 10 days $= \frac{8\times 5}{45} = \frac{8}{9}$
Remaining work $= 1-\frac{8}{9} = \frac{1}{9}$
Now it is turn of ‘A’ for the eleventh day.
∴ Time taken by ‘A’ in doing $\frac{1}{9}$ work $= \frac{1}{9}\times 9 = 1\ day$
$\therefore Required\ Time\ = 10+1=11\ days$
Hence option [C] is correct answer.