$12$ men complete a work in $9$ days. After they have worked for $6$ days, $6$ more men join them. How many days will they take to complete the remaining work?

The correct option is $A.2$ days

$1$ man’s $1$ day’s work = $\frac{1}{108}$

$12$ men’s $6$ days’ work = $(\frac{1}{9}\times 6)=\frac{2}{3}$

$\therefore$ Remaining work = $(1-\frac{2}{3})=\frac{1}{3}$

$18$ men’s $1$ days work $=$ $(\frac{1}{108}\times 18)=\frac{1}{6}$

$\frac{1}{6}$ work is done by them in $1$ day.

$\frac{1}{3}$ work is done by them in$(6\times \frac{1}{3})$ $=2$ days

Therefore, remaining work will be completed in $2$ days.