A and B can complete a piece of work in 12 an | vedclass

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(A) 's $1 	ext{ day}$'s work $= \frac{1}{12}$.$B$'s $1 	ext{ day}$'s work $= \frac{1}{18}$.Part of work done by $A$ and $B$ in the first $2 	ext{ d

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$14\frac{1}{3}$

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(A) 's $1 	ext{ day}$'s work $= \frac{1}{12}$.$B$'s $1 	ext{ day}$'s work $= \frac{1}{18}$.Part of work done by $A$ and $B$ in the first $2 	ext{ days} = \frac{1}{12} + \frac{1}{18} = \frac{3+2}{36} = \frac{5}{36}$.In $14 	ext{ days}$ ($7$ cycles of $2 	ext{ days}$),the work done $= 7 	imes \frac{5}{36} = \frac{35}{36}$.Remaining work $= 1 - \frac{35}{36} = \frac{1}{36}$.On the $15^{	ext{th}}$ day,$A$ works. Since $A$ completes $\frac{1}{12}$ work in $1 	ext{ day}$,$A$ will complete $\frac{1}{36}$ work in $\frac{1}{36} 	imes 12 = \frac{1}{3} 	ext{ day}$.Total time taken $= 14 + \frac{1}{3} = 14\frac{1}{3} 	ext{ days}$.

$A$ and $B$ can complete a piece of work in $12$ and $18 \text{ days}$ respectively. $A$ begins to do the work and they work alternatively one at a time for $1 \text{ day}$ each. The whole work will be completed in (in $\text{days}$):

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