Question

A tank is emptied by $2$ pipes and filled by a third. If the 1st two can empty the tank in $2$ and $3$ hrs, respectively and third can fill it in $4$ hours, how much time will it take to empty the tank ${\left(\frac{4}{5}\right)}^{th}$ fully, when all three are open?

Answer 1

Let the time taken to completely empty the tank is $x$ hrs.

$\Rightarrow \frac{1}{x}=\frac{1}{2}+\frac{1}{3}+(-\frac{1}{4})$
$\Rightarrow \frac{1}{x}=\frac{7}{12}$

$\Rightarrow x=\frac{12}{7}hrs$
Complete tank will be emptied in $\frac{12}{7}$ hrs
$\therefore {\left(\frac{4}{5}\right)}^{th}$ tank will be emptied in = $\frac{12}{7}\times \frac{4}{5}=\frac{48}{35}$ hrs