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(C) Let the work done by $A$ in $1$ day be $a$,by $B$ be $b$,and by $C$ be $c$.Given that $A$ does the work in the same time as $B$ and $C$ together,s

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$25$

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(C) Let the work done by $A$ in $1$ day be $a$,by $B$ be $b$,and by $C$ be $c$.Given that $A$ does the work in the same time as $B$ and $C$ together,so $a = b + c$.Given $(A+B)$'s $1$ day work is $\frac{1}{10}$,so $a + b = \frac{1}{10}$.Given $C$'s $1$ day work is $c = \frac{1}{50}$.Since $a = b + c$,we substitute $b = a - c$ into the equation $a + b = \frac{1}{10}$:$a + (a - c) = \frac{1}{10}$$2a - c = \frac{1}{10}$$2a - \frac{1}{50} = \frac{1}{10}$$2a = \frac{1}{10} + \frac{1}{50} = \frac{5+1}{50} = \frac{6}{50} = \frac{3}{25}$$a = \frac{3}{50}$.Now,find $b$ using $a + b = \frac{1}{10}$:$b = \frac{1}{10} - a = \frac{1}{10} - \frac{3}{50} = \frac{5-3}{50} = \frac{2}{50} = \frac{1}{25}$.Thus,$B$ alone can complete the work in $25$ days.

$A$ can do a certain work in the same time in which $B$ and $C$ together can do it. If $A$ and $B$ together could do it in $10$ days and $C$ alone in $50$ days,then $B$ alone could do the work in (in days)

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