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Hint: We want to find the time taken by each tap. For that, assume that the tap $1$ takes "$x$ hours" of time and the tap $2$ takes "$(x+3)$ hours" of time. After that, take the time of two taps for one hour. Arrange the problem as mentioned in question, you will get the answer.

Complete step-by-step answer:

Two taps can fill a tank in $3\dfrac{1}{13}$ hours. One tap takes $3$ hours more than the other tap to fill the tank. Let, tap $1$ take "$x$ hours" of time, then tap $2$ takes "$(x+3)$ hours" of time.

In one hour tap $1$ does $\dfrac{1}{x}$ of work, i.e., fill $\dfrac{1}{x}$of the tank.

In one hour tap $2$, fills $\dfrac{1}{x+3}$ of the tank.

So work done by both the taps in one hour is sum of the work done in one hour by tap $1$and tap $2$ i.e. $\left( \dfrac{1}{x}+\dfrac{1}{x+3} \right)$.

So for $\left( 3\dfrac{1}{13} \right)$hours work done, the entire tank will be filled, as is given, or,

$\left( \dfrac{1}{x}+\dfrac{1}{x+3} \right)\left( 3\dfrac{1}{13} \right)=1$

So simplifying we get,

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

$\begin{align}

& \left( \dfrac{1}{x}+\dfrac{1}{x+3} \right)\left( \dfrac{40}{13} \right)=1 \\

& 80x+120=13{{x}^{2}}+39x \\

& 13{{x}^{2}}-41x-120=0 \\

\end{align}$

We have got a quadratic equation.

Now we know,

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

So here, $a=13,$ $b=-41$, $c=-120$.

Now,

$\begin{align}

& x=\dfrac{-(-41)\pm \sqrt{{{(-41)}^{2}}-4(13)(-120)}}{2(13)} \\

& x=\dfrac{41\pm \sqrt{1681+6240}}{26} \\

& x=\dfrac{41\pm \sqrt{7921}}{26}=\dfrac{41\pm 89}{26} \\

& x=\dfrac{41\pm 89}{26} \\

\end{align}$

So we get,

$x=5$ and $x=-\dfrac{24}{13}$.

Time cannot be negative so we are going to reject the negative value of $x$.

So we get the final answer as $x=5$hours.

Hence, the time taken by tap $1$ to fill the tank is $5$hours and that of tap $2$ is $x+3=5+3=8$hours.

Note: Read the question carefully. Also, take care that no term is missing. Your concept regarding this problem should be clear. Do not make silly mistakes. While simplifying take utmost care that no signs are missing. Solve the problem in step by step way.

Complete step-by-step answer:

Two taps can fill a tank in $3\dfrac{1}{13}$ hours. One tap takes $3$ hours more than the other tap to fill the tank. Let, tap $1$ take "$x$ hours" of time, then tap $2$ takes "$(x+3)$ hours" of time.

In one hour tap $1$ does $\dfrac{1}{x}$ of work, i.e., fill $\dfrac{1}{x}$of the tank.

In one hour tap $2$, fills $\dfrac{1}{x+3}$ of the tank.

So work done by both the taps in one hour is sum of the work done in one hour by tap $1$and tap $2$ i.e. $\left( \dfrac{1}{x}+\dfrac{1}{x+3} \right)$.

So for $\left( 3\dfrac{1}{13} \right)$hours work done, the entire tank will be filled, as is given, or,

$\left( \dfrac{1}{x}+\dfrac{1}{x+3} \right)\left( 3\dfrac{1}{13} \right)=1$

So simplifying we get,

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

$\begin{align}

& \left( \dfrac{1}{x}+\dfrac{1}{x+3} \right)\left( \dfrac{40}{13} \right)=1 \\

& 80x+120=13{{x}^{2}}+39x \\

& 13{{x}^{2}}-41x-120=0 \\

\end{align}$

We have got a quadratic equation.

Now we know,

$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

So here, $a=13,$ $b=-41$, $c=-120$.

Now,

$\begin{align}

& x=\dfrac{-(-41)\pm \sqrt{{{(-41)}^{2}}-4(13)(-120)}}{2(13)} \\

& x=\dfrac{41\pm \sqrt{1681+6240}}{26} \\

& x=\dfrac{41\pm \sqrt{7921}}{26}=\dfrac{41\pm 89}{26} \\

& x=\dfrac{41\pm 89}{26} \\

\end{align}$

So we get,

$x=5$ and $x=-\dfrac{24}{13}$.

Time cannot be negative so we are going to reject the negative value of $x$.

So we get the final answer as $x=5$hours.

Hence, the time taken by tap $1$ to fill the tank is $5$hours and that of tap $2$ is $x+3=5+3=8$hours.

Note: Read the question carefully. Also, take care that no term is missing. Your concept regarding this problem should be clear. Do not make silly mistakes. While simplifying take utmost care that no signs are missing. Solve the problem in step by step way.