Question

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

Answer 1

Let the time taken by faster pipe to fill the cistern be $x$ minutes

Therefore, time taken by slower pipe to fill the cistern $=(x+3)\text{minutes}$

Since the faster pipe takes $x$ minutes to fill the cistern.

Portion of cistern filled by the faster pipe in $1$ minute $=\frac{1}{x}$

Portion of cistern filled by the faster pipe in $1$ minute $=\frac{1}{x+3}$

Portion of cistern filled by the two pipes in $1$ minute $=\frac{1}{\left(\frac{40}{13}\right)}=\frac{13}{40}$

Now according to the question-

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

$\frac{x+3+x}{x(x+3)}=\frac{13}{40}$

$40(2x+3)=13x(x+3)$

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

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

Here,

$a=13,b=-41,c=-120$

Now by quadratic formula-

$x=\frac{-(-41)\pm \sqrt{{(-41)}^2-4\times \left(13\right)\times (-120)}}{2\times 13}$

$\Rightarrow x=\frac{41\pm \sqrt{1681+6240}}{26}$

$\Rightarrow x=\frac{41\pm \sqrt{7921}}{26}$

$\Rightarrow x=\frac{41\pm 89}{26}$

$\because x\ne -ve$

$\therefore x=\frac{41+89}{26}=5$

Therefore,

Time taken by faster pipe to fill the cistern $=5\text{minutes}$

Time taken by slower pipe to fill the cistern $=(5+3)=8\text{minutes}$