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(B) Suppose $C$ alone can do this work in $x$ days.Therefore,$C$ will do $\frac{1}{x}$ work in $1$ day.Now,work done by $(B + C)$ in $1$ day $= \frac{

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) Suppose $C$ alone can do this work in $x$ days.Therefore,$C$ will do $\frac{1}{x}$ work in $1$ day.Now,work done by $(B + C)$ in $1$ day $= \frac{1}{16}$.Therefore,work done by $B$ in $1$ day $= \left(\frac{1}{16} - \frac{1}{x}\right)$.And,work done by $(A + B)$ in $1$ day $= \frac{1}{12}$.Therefore,work done by $A$ in $1$ day $= \frac{1}{12} - \left(\frac{1}{16} - \frac{1}{x}\right) = \frac{4-3}{48} + \frac{1}{x} = \frac{1}{48} + \frac{1}{x}$.As per the question,work done by $A$ in $5$ days $+$ work done by $B$ in $7$ days $+$ work done by $C$ in $13$ days $= 1$ (whole work).Therefore,$5\left(\frac{1}{48} + \frac{1}{x}\right) + 7\left(\frac{1}{16} - \frac{1}{x}\right) + \frac{13}{x} = 1$.$\frac{5}{48} + \frac{5}{x} + \frac{7}{16} - \frac{7}{x} + \frac{13}{x} = 1$.$\frac{5}{48} + \frac{21}{48} + \frac{11}{x} = 1$.$\frac{26}{48} + \frac{11}{x} = 1$.$\frac{11}{x} = 1 - \frac{13}{24} = \frac{11}{24}$.Thus,$x = 24$.$C$ alone would complete this work in $24$ days.

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