$A$ can do a work in $18$ days, $B$ in $9$ days and $C$ in $6$ days. $A$ and $B$ start working together and after $2$ days $C$ joins them. What is the total number of days taken to finish the work?

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Solution

$A$ does $1$ unit of work in $18$ days
$∴$ he does $\frac{1}{18}$ units of work in $1$ day

$B$ does $1$ unit of work in $9$ days
$∴$ he does $\frac{1}{9}$ units of work in $1$ day

$C$ does $1$ unit of work in $6$ days
$∴$ he does $\frac{1}{6}$ units of work in $1$ day

$∴$ in two days, work done by $A$ and $B$
$= 2 (\frac{1}{18} + \frac{1}{9}) = \frac{1}{3}$

Work left $= 1 - \frac{1}{3} = \frac{2}{3}$

Work done by $A, B$ and $C$ in one day
$= \frac{1}{18} + \frac{1}{9} + \frac{1}{6} = \frac{1}{3}$

Days required for $A, B, C$ to do $\frac{2}{3}$ work
$= \frac{2}{3} \div \frac{1}{3} = 2$ days

Correct answer is 4 days

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