A can do a piece of work in 30 days,B in 50 d | vedclass

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(A) $(A+B)$'s $1$ day's work $= \frac{1}{30} + \frac{1}{50} = \frac{5+3}{150} = \frac{8}{150} = \frac{4}{75}$ units.$(A+C)$'s $1$ day's work $= \frac{

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$17 \frac{32}{35}$

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(A) $(A+B)$'s $1$ day's work $= \frac{1}{30} + \frac{1}{50} = \frac{5+3}{150} = \frac{8}{150} = \frac{4}{75}$ units.$(A+C)$'s $1$ day's work $= \frac{1}{30} + \frac{1}{40} = \frac{4+3}{120} = \frac{7}{120}$ units.Work done in a cycle of $2$ days $= \frac{4}{75} + \frac{7}{120} = \frac{32+35}{600} = \frac{67}{600}$ units.In $8$ cycles ($16$ days),work done $= 8 	imes \frac{67}{600} = \frac{67}{75}$ units.Remaining work $= 1 - \frac{67}{75} = \frac{8}{75}$ units.On the $17^{th}$ day,$(A+B)$ works. They do $\frac{4}{75}$ units. Remaining work $= \frac{8}{75} - \frac{4}{75} = \frac{4}{75}$ units.On the $18^{th}$ day,$(A+C)$ works. They do $\frac{7}{120}$ units per day. Time taken to finish remaining work $= \frac{4/75}{7/120} = \frac{4}{75} 	imes \frac{120}{7} = \frac{4 	imes 8}{5 	imes 7} = \frac{32}{35}$ days.Total time $= 17 + \frac{32}{35} = 17 \frac{32}{35}$ days.

$A$ can do a piece of work in $30$ days,$B$ in $50$ days,and $C$ in $40$ days. If $A$ is assisted by $B$ on one day and by $C$ on the next day alternately,the work will be completed in (in days):

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