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(C) Work done by $A$ in $1$ day $= \frac{1}{9}$.Work done by $B$ in $1$ day $= \frac{1}{12}$.Work done in a cycle of $2$ days (starting with $A$) $= \

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$10 \frac{1}{4}$

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(C) Work done by $A$ in $1$ day $= \frac{1}{9}$.Work done by $B$ in $1$ day $= \frac{1}{12}$.Work done in a cycle of $2$ days (starting with $A$) $= \frac{1}{9} + \frac{1}{12} = \frac{4+3}{36} = \frac{7}{36}$.In $10$ days ($5$ cycles),work done $= 5 	imes \frac{7}{36} = \frac{35}{36}$.Remaining work $= 1 - \frac{35}{36} = \frac{1}{36}$.On the $11$th day,it is $A$'s turn. $A$ does $\frac{1}{9}$ work in $1$ day.Time taken by $A$ to complete $\frac{1}{36}$ work $= \frac{1/36}{1/9} = \frac{9}{36} = \frac{1}{4}$ day.Total time $= 10 + \frac{1}{4} = 10 \frac{1}{4}$ days.

$A$ and $B$ working separately can do a piece of work in $9$ and $12$ days,respectively. If they work for a day alternatively with $A$ beginning,the work would be completed in (in days):

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