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# Two pipes can fill a tank in 12 and 20 hours respectively. The pipes are open simultaneously and it is found that due to leakage in the bottom, 30 minutes extra are taken for the cistern to be filled up. If the cistern is full, in what time would the leak empty it?

Last updated date: 20th Jun 2024
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Hint: We solve this problem by considering the height of cistern as $'H'$ and the time taken by the leakage to empty it as $'x'$ then we use the given condition that when both pipes are open it takes 30 minutes extra to fill the tank due to leakage to find the required value. While applying the conditions we need to keep in mind that times of individual taps should not be added to get time for filling tanks together. We need to add the height not time.

Let us assume that the height of cistern as $'H'$
We are given that the first pipe takes 12 hours to fill the tank.
So, we can write that the first pipe fills $'\dfrac{H}{12}'$ height of the cistern in 1 hour.
We are given that the second pipe takes 20 hours to fill the tank.
So, we can write that the second pipe fills $'\dfrac{H}{20}'$ height of the tank in 1 hour.
Let us assume that the height of tank filled with both the pipes simultaneously as $'h'$ then we get
$\Rightarrow h=\dfrac{H}{12}+\dfrac{H}{20}$
Let us assume that both pipes together take $'t'$ time to fill the tank without leakage then we get
\begin{align} & \Rightarrow t=\dfrac{H}{\dfrac{H}{12}+\dfrac{H}{20}} \\ & \Rightarrow t=\dfrac{12\times 20}{12+20} \\ & \Rightarrow t=\dfrac{15}{2}hrs \\ \end{align}
Here, we can say that it takes $\dfrac{15}{2}$ hours to fill the tank by both pipes together.
Let us assume that the time taken to fill the tank when there is leakage to empty the tank as $'x'$
We are given that when both pipes are open it takes 30 minutes extra to fill the tank due to leakage
We know that the conversion of time that is
$1hr=60\min$
By using this conversion we get
$\Rightarrow 30\min =\dfrac{1}{2}hr$
Now, by converting the given condition into mathematical equation we get
$\Rightarrow t+\dfrac{1}{2}=x$
Now, by substituting the value of $'t'$ in above equation we get
\begin{align} & \Rightarrow x=\dfrac{15}{2}+\dfrac{1}{2} \\ & \Rightarrow x=8hr \\ \end{align}
Therefore we can say that it takes 8 hours to fill the tank when there is leakage.
So, we can say that both pipes combined with leakage empties $'\dfrac{H}{8}'$ height of the tank in 1 hour.
Let us assume that the leakage takes $'y'$ hours to empty the tank.
So, we can say that the leakage empties $'\dfrac{H}{y}'$ height in one hour.
Now, we can say that the height filled by both pipes to fill the tank when there is leakage as
\begin{align} & \Rightarrow \dfrac{H}{12}+\dfrac{H}{20}-\dfrac{H}{y}=\dfrac{H}{8} \\ & \Rightarrow \dfrac{1}{y}=\dfrac{1}{12}+\dfrac{1}{20}-\dfrac{1}{8} \\ \end{align}
Now by adding the terms using the LCM we get
\begin{align} & \Rightarrow \dfrac{1}{y}=\dfrac{10+6-15}{120} \\ & \Rightarrow y=120 \\ \end{align}
Therefore, we can say that the leakage takes 120 hours to empty the tank.

Note: Students may make mistakes in taking the time of both pipes together to fill the tank.
We have the time taken by both pipes to fill the tank as
$\Rightarrow t=\dfrac{H}{\dfrac{H}{12}+\dfrac{H}{20}}$
Bu students may do mistake and add the times directly to get the time taken by both pipes together that is
$\Rightarrow t=12+20$
This gives the wrong answer because we are not allowed to add the times when the situations are taken together. But instead, we can add work done in small change in time as we considered 1 hour in the solution.