Question

Two pipes running together can fill a cistern in $3\frac{1}{13}$ minutes. If one pipe takes $3$ minutes more than the other to fill it, find the time in which each pipe would fill cistern.

Answer 1

Let the time taken by the faster pipe to fill the cistern is $x$ minutes, then the time taken the slower pipe will be $\left(x+3\right)$ minutes.

The portion of cistern filled by the faster pipe in 1 minute is $\frac{1}{x}$.

The portion of cistern filled by the faster pipe in $\frac{40}{13}$ minutes is $\frac{40}{13x}$.

The portion of cistern filled by the slower pipe in $\frac{40}{13}$ minutes is $\frac{40}{13\left(x+3\right)}$.

As the cistern is filled fully in $\frac{40}{13}$ minutes, then,

$\Rightarrow \frac{40}{13x}+\frac{40}{13\left(x+3\right)}=1$

$\Rightarrow \frac{1}{x}+\frac{1}{\left(x+3\right)}=\frac{13}{40}$

$\Rightarrow 40\left(2x+3\right)=13\left(x+3\right)$

$\Rightarrow 13x^2-41x-120=0$

$\Rightarrow 13x^2-65x+24x-120=0$

$\Rightarrow \left(x-5\right)\left(13x+24\right)=0$

$\therefore x=5,x=-\frac{24}{13}$

$\because$ $x$ can not be a negative

Therefore, the faster pipe will fill the cistern in 5 minutes and the slower pipe will fill the cistern in $5+3=8$ minutes.