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

Hint: In this question, we will take time taken by one of the pipes to be variable, then find the fraction of cistern filled in 1 minute by each pipe and use it to find expression for time taken by both pipes together to fill the cistern. We will equate this with data in question to find the solution.

Complete step-by-step answer:
We are given that one pipe takes 3 minutes more than the other to fill the cistern. Let the time taken by one of the pipes to fill the cistern alone be $x$ minutes. Then, the time taken by others to fill the cistern alone will be 3 minutes more than this, that is \[3+x\] minutes.
Therefore, fraction of the cistern filled by first pipe in one minute is given by $\dfrac{1}{x}$
And the time taken by the other pipe to fill the cistern in one minute is given by $\dfrac{1}{x+3}$
Therefore, if both pipes are together to fill the same cistern, then fraction of cistern filled by them in one minute will be equal to $\dfrac{1}{x}+\dfrac{1}{x+3}$
Taking LCM we get,
$\begin{align}
  & \dfrac{x+3}{x\left( x+3 \right)}+\dfrac{x}{x\left( x+3 \right)} \\
 & =\dfrac{x+3+x}{x\left( x+3 \right)} \\
 & =\dfrac{3+2x}{x\left( x+3 \right)} \\
\end{align}$
Therefore, the time taken by both the pipes together to completely fill the cistern will be $=\dfrac{x\left( x+3 \right)}{3+2x}$
Now, according to question, time taken by both pipes together to fill the cistern is $=3\dfrac{1}{13}=\dfrac{40}{13}$
Therefore,
$\dfrac{x\left( x+3 \right)}{3+2x}=\dfrac{40}{13}$
Cross multiplying, we get,
$13x\left( x+3 \right)=40\left( 3+2x \right)$
Applying distributive law, we get,

$13{{x}^{2}}+39x=120+80x$
Subtracting $80x+120$ from both sides, we get,
$\begin{align}
  & 13{{x}^{2}}+39x-80x-120=0 \\
 & \Rightarrow 13{{x}^{2}}-41x-120=0 \\
\end{align}$
Applying quadratic equation formula, we get,
$\begin{align}
  & x=\dfrac{-\left( -41 \right)\pm \sqrt{{{41}^{2}}-4\times 13\times \left( -120 \right)}}{2\times 13} \\
 & =\dfrac{41\pm \sqrt{1681+4\times 13\times 120}}{26} \\
 & =\dfrac{41\pm \sqrt{1681+6240}}{26} \\
 & =\dfrac{41\pm \sqrt{7921}}{26} \\
\end{align}$
Writing value of root, we get,
$=\dfrac{41\pm 89}{26}$
Separating plus and minus sign, we get,
$x=\dfrac{41-89}{26}$ or $x=\dfrac{41+89}{26}$
$\Rightarrow x=\dfrac{-48}{26}$ or $x=\dfrac{130}{26}$
Since, time cannot be negative,
Therefore,$x=\dfrac{130}{26}=5$
Hence, time taken by one of the pipes to fill the cistern alone will be 5 minutes.
And time taken by the other pipe to fill the cistern will be 5+3 = 8 minutes.

Note: In this type of questions, we use the concept of inverse proportionality and first find the fraction of cistern filled in one minute. Then reverse the value to find time taken.