Question

Two pipes running together can fill a cistern in $\frac{40}{13}$ minutes. If one pipe takes $3$ minutes more than the other to fill it, find the time (in min) taken by the faster pipe to fill the cistern.

Answer 1

Let the time taken by the faster pipe be $x$ mins. The time taken by the other pipe will be $(x+3)$ mins.

The two pipes can together fill the cistern in $\frac{40}{13}$ mins.

$\therefore \frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}$

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

$\Rightarrow 80x+120=13x^2+39x$

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

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

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

Since time cannot be negative, we take only the positive solution, i.e, $x=5$ mins.