Three pipes X, Y and Z can fill a tank in $6$ hours. After working at it together for $2$ hours, $Z$ is closed and $X$ and $Y$ can fill the remaining part in $7$ hours. The number of hours taken by $Z$ alone to fill the tank is:

A) 10

B) 12

C) 14

D) 16


Option C


Part filled in $2$ hours  $=\frac{2}{6}$  $=\frac{1}{3}$

Remaining part  $=\left(1-\frac{1}{3}\right)$  $=\frac{2}{3}$.

$\therefore$  $(X+Y)$’s $7$  hour’s work $=\frac{2}{3}$.

$(X+Y)$’s $1$ hour’s work $=\frac{2}{21}$.

$\therefore$ $Z$’s $1$ hour’s work = $\{(X+Y+Z)$’s $1$ hour’s work$\}$  - $\{(X+Y)$’s $1$ hour’s work$\}$

$=\left(\frac{1}{6}-\frac{2}{21}\right)$ $=\frac{1}{14}$

$\therefore$ $Z$ alone can fill the tank in 14 hours