Question

A tap fills a tank in 28 hours and an outlet can empty the full tank in 35 hours. In how many hours will the tank be filled, if both the tap and outlet are opened simultaneously?

Answer 1

Hint: First consider the height of the tank as ‘H’. Here, the tap is trying to fill the tank while the outlet is trying to empty the tank so there will be some change in height (\[dh\]) effectively within a small change in time (\[dt\]). Since we cannot add or subtract the time in these cases we have to take a small change in time and find the height it will increase effectively due to both tap and outlet and then find the total time required to fill the tank if in a small amount of time there is some effective change in height by integrating by applying the limits as
Limits of ‘H’ changes from 0 to H.
Limits of time changes from 0 to T.

Complete step-by-step solution
Let us consider the height of the tank as ‘H’
Also, let us consider the small amount of time ‘\[dt\]’ hours.
By considering the tap that fills the tank in 28 hours, let us find how much height it will fill the tank in the time \[dt\] as follows
\[{{h}_{1}}=\dfrac{H}{28}.dt\]
Now, by considering the outlet that empty the tank in 35 hours, let us find how much height it will empty the tank in the time \[dt\] as follows
\[{{h}_{2}}=\dfrac{H}{35}.dt\]
Now let us find the effective height \[dh\]that the tank will be filled due to both tap and outlet as follows
\[\begin{align}
  & \Rightarrow dh={{h}_{1}}-{{h}_{2}} \\
 & \Rightarrow dh=\dfrac{H}{28}.dt-\dfrac{H}{35}.dt \\
\end{align}\]
Taking out common \[H.dt\]and subtracting we get
\[\begin{align}
  & \Rightarrow dh=H.dt\left( \dfrac{1}{28}-\dfrac{1}{35} \right) \\
 & \Rightarrow dh=H.dt\left( \dfrac{7}{28\times 35} \right) \\
 & \Rightarrow dh=\dfrac{H.dt}{140} \\
\end{align}\]
Now let us integrate on both sides applying the limits as ‘H’ changes from 0 to H and limits of time changes from 0 to T (where T is total time required to full the tank) we will get
\[\begin{align}
  & \Rightarrow \int\limits_{0}^{H}{dh}=\dfrac{H}{140}\int\limits_{0}^{T}{dt} \\
 & \Rightarrow H-0=\dfrac{H}{140}\left( T-0 \right) \\
 & \Rightarrow T=140 \\
\end{align}\]
Therefore it requires 140 hours to full the tank.

Note: Many students will take the time as the changing parameter that is responsible for the change in height and gives the answer by subtracting the two times given. But the parameter that is changing is height in small change in time. So, we need to find the height changed due to the small changes in height and solve accordingly.