Question

How much time will the leak take to empty the full cistern?
I. The cistern is normally filled in 9 hours.
II. It takes one hour more than the usual time to fill the cistern because of the leak in the bottom.
(a) I alone sufficient while II alone is not sufficient to answer
(b) II alone sufficient while I alone is not sufficient to answer
(c) Either I or II alone sufficient to answer
(d) Both I and II are not sufficient to answer
(e) Both I and II are sufficient to answer

Answer 1

Hint:
Here, we need to find the time taken by the leak to empty the full cistern. Using condition I, we will find the volume of water filled in 1 hour without the leak. Using condition II, we will find the volume of water filled in 1 hour with the leak. Then we will find the difference of the volumes obtained Using this, we will find the required number of hours taken by the leak to empty the full cistern.

Complete step by step solution:
Let the capacity/volume of the cistern be \[x\] cubic units.
This means that the volume of water needed to fill the cistern is \[x\] cubic units.
I. We know that the cistern is filled in 9 hours.
Therefore, we get
Volume of water filled in 9 hours (without a leak on the bottom) \[ = \]\[x\] cubic units.
Dividing by 9, we get
Volume of water filled in 1 hour (without a leak on the bottom) \[ = \]\[\dfrac{x}{9}\] cubic units.
II. It is given that it takes one hour more than the regular time to fill the cistern because of the leak present at the bottom.
This means that it takes 10 hours to fill the cistern with the leak on the bottom.
Volume of water filled in 10 hours (with a leak on the bottom) \[ = \]\[x\] cubic units.
Therefore, we get
Volume of water filled in 1 hour (with a leak on the bottom) \[ = \]\[\dfrac{x}{{10}}\] cubic units.
Now, the rate of filling the cistern remains the same. The time taken is 1 hour more because the leak removes some water.
Therefore, the volume of water filled in 1 hour without the leak minus the volume of water removed due to the leak in 1 hour is equal to the volume of water filled in 1 hour with the leak.
Thus, we can form the equation
\[\dfrac{x}{9} - {\text{Volume of water leaked in 1 hour}} = \dfrac{x}{{10}}\]
Rewriting the equation, we get
\[ \Rightarrow \]Volume of water leaked in 1 hour\[ = \dfrac{x}{9} - \dfrac{x}{{10}}\]
Taking the L.C.M., we get
\[ \Rightarrow \]Volume of water leaked in 1 hour\[ = \dfrac{{10x - 9x}}{{90}} = \dfrac{x}{{90}}\] cubic units
Let the number of hours it takes for the leak to empty the cistern be \[y\] hours.
Now, the volume of cistern taken out by the leak in \[y\] hours is equal to the volume of water leaked in 1 hour, multiplied by the number of hours.
The volume of cistern taken out in \[y\] hours is \[x\] cubic units.
Therefore, we get
\[ \Rightarrow x = \dfrac{x}{{90}} \times y\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 1 = \dfrac{1}{{90}} \times y\\ \Rightarrow y = 90\end{array}\]
Therefore, the number of hours it takes for the leak to empty the cistern is 90 hours.
Thus, we need both I and II to find the required number of hours.

\[\therefore \] The correct option is option (e).

Note:
Here, condition I alone is not sufficient to answer since it only tells how much time it usually takes to fill the cistern. There is no information about the rate of flow of water from the leak. In II, we do not know what the ‘usual time’ is without using statement I. Therefore, both I and II are needed to find the required answer.
We have taken the capacity/volume of the cistern as \[x\] cubic units. Another way to solve this without \[x\] is by letting the work done to fill the cistern be 1.