A can do a piece of work in 120 days and B ca | vedclass

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(C) 's $1$ day's work $= \frac{1}{120}$.$B$'s $1$ day's work $= \frac{1}{150}$.$(A+B)$'s $1$ day's work $= \frac{1}{120} + \frac{1}{150} = \frac{5+4}{

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$240$

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(C) 's $1$ day's work $= \frac{1}{120}$.$B$'s $1$ day's work $= \frac{1}{150}$.$(A+B)$'s $1$ day's work $= \frac{1}{120} + \frac{1}{150} = \frac{5+4}{600} = \frac{9}{600} = \frac{3}{200}$.$(A+B)$ work together for $20$ days,so work done $= 20 	imes \frac{3}{200} = \frac{3}{10}$.After $B$ leaves,$A$ works alone for $12$ days. $A$'s $12$ day's work $= 12 	imes \frac{1}{120} = \frac{1}{10}$.Then $C$ joins $A$ and they work for $48$ days to complete the remaining work.$A$'s work during these $48$ days $= 48 	imes \frac{1}{120} = \frac{2}{5}$.Total work done by $A$ and $B$ before $C$ joined $= \frac{3}{10} + \frac{1}{10} = \frac{4}{10} = \frac{2}{5}$.Total work done by $A$ in the final $48$ days $= \frac{2}{5}$.Total work done by $A$ and $B$ throughout $= \frac{3}{10} + \frac{1}{10} + \frac{2}{5} = \frac{4}{10} + \frac{4}{10} = \frac{8}{10} = \frac{4}{5}$.Remaining work $= 1 - \frac{4}{5} = \frac{1}{5}$.This remaining work is done by $C$ in $48$ days.$C$'s $1$ day's work $= \frac{1/5}{48} = \frac{1}{240}$.Therefore,$C$ alone can finish the work in $240$ days.

$A$ can do a piece of work in $120$ days and $B$ can do it in $150$ days. They work together for $20$ days. Then $B$ leaves and $A$ alone continues the work. After $12$ days,$C$ joins $A$ and the work is completed in $48$ days more. In how many days can $C$ do it if he works alone?

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