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(C) Work done by $A$ in $1$ day $= 1/9$.Work done by $B$ in $1$ day $= 1/12$.Work done by $A$ and $B$ together in $2$ days (alternating) $= 1/9 + 1/12

Vedclass · Accepted Answer

$10 \frac{1}{4}$

Vedclass · Answer

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

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

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