A, B and C can do a piece of work in 16, 32 a | vedclass

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(D) The $1$ day's work of $A, B$ and $C$ is $\frac{1}{16}, \frac{1}{32}$ and $\frac{1}{48}$ respectively.$(A+B+C)$'s $1$ day's work $= \frac{1}{16} +

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$10\frac{4}{9}$

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(D) The $1$ day's work of $A, B$ and $C$ is $\frac{1}{16}, \frac{1}{32}$ and $\frac{1}{48}$ respectively.$(A+B+C)$'s $1$ day's work $= \frac{1}{16} + \frac{1}{32} + \frac{1}{48} = \frac{6+3+2}{96} = \frac{11}{96}$.Work done by $A, B$ and $C$ in the first $4$ days $= 4 	imes \frac{11}{96} = \frac{11}{24}$.Remaining work $= 1 - \frac{11}{24} = \frac{13}{24}$.Let the total time taken be $x$ days. $B$ left $2$ days before completion,so $B$ worked for $(x-2)$ days. $A$ worked for the entire $x$ days. $C$ worked for $4$ days.Since $C$ left after $4$ days,the remaining work $\frac{13}{24}$ was done by $A$ and $B$.Work done by $A$ in $x$ days + Work done by $B$ in $(x-2)$ days $= \frac{13}{24}$.$\frac{x}{16} + \frac{x-2}{32} = \frac{13}{24}$.Multiply by $96$: $6x + 3(x-2) = 52$.$6x + 3x - 6 = 52 \Rightarrow 9x = 58 \Rightarrow x = \frac{58}{9} = 6\frac{4}{9}$.Total days $= 4$ (initial) $+ 6\frac{4}{9} = 10\frac{4}{9}$ days.

$A, B$ and $C$ can do a piece of work in $16, 32$ and $48$ days,respectively. They started working together,but $C$ left after working $4$ days and $B$ left $2$ days before the completion of the work. The total number of days taken for the completion of the work was (in days):

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