Understanding Tank Filling and Emptying Problems

This problem involves calculating the time taken by a leak to empty a tank, given the time taken by a pump to fill it alone and the time taken by the pump to fill it when the leak is active. We can approach this by considering the rate at which work is done (filling or emptying the tank).

Calculating Individual Rates

Let the total capacity of the tank be 1 unit (representing a full tank). The rate of filling or emptying is the amount of tank filled or emptied per hour.

• The pump can fill the tank in 4 hours. This means the pump fills $\frac{1}{4}$ of the tank per hour. We can call this the filling rate of the pump, $R_p$. So, $R_p = \frac{1}{4}$ tank/hour.
• When the leak is present, the tank is filled in 5 hours. This means the net filling rate (pump filling minus leak emptying) is $\frac{1}{5}$ of the tank per hour. We can call the emptying rate of the leak $R_l$. The combined rate is $R_p - R_l$. So, $R_p - R_l = \frac{1}{5}$ tank/hour.

Finding the Leakage Rate

We know the pump's filling rate ($R_p = \frac{1}{4}$) and the combined rate ($R_p - R_l = \frac{1}{5}$). We can use these to find the leak's emptying rate ($R_l$).

Substitute the value of $R_p$ into the combined rate equation:

\( \frac{1}{4} - R_l = \frac{1}{5} \)

Now, we solve for $R_l$:

\( R_l = \frac{1}{4} - \frac{1}{5} \)

To subtract these fractions, we find a common denominator, which is 20:

\( R_l = \frac{1 \times 5}{4 \times 5} - \frac{1 \times 4}{5 \times 4} \)

\( R_l = \frac{5}{20} - \frac{4}{20} \)

\( R_l = \frac{5 - 4}{20} \)

\( R_l = \frac{1}{20} \)

The leak empties $\frac{1}{20}$ of the tank per hour. This is the emptying rate of the leak.

Calculating Time for Leak to Empty Tank

The time taken to empty the full tank by the leak alone is the reciprocal of its emptying rate.

Time = $\frac{\text{Total Capacity}}{\text{Rate}}$

Since the total capacity is 1 unit and the leak's rate is $R_l = \frac{1}{20}$, the time taken for the leak to empty the tank ($T_l$) is:

\( T_l = \frac{1}{R_l} = \frac{1}{\frac{1}{20}} \)

\( T_l = 1 \times 20 = 20 \)

So, it will take the leakage 20 hours to empty the tank when it is full.

Summary of Rates and Times

The leak alone empties the tank at a rate of $\frac{1}{20}$ tank per hour. Therefore, it will take 20 hours to empty the entire tank.

Revision Table: Tank and Leak Problems

Additional Information: Working with Rates

Problems involving filling and emptying tanks (or pipes and cisterns) are often solved using the concept of work rates. The key idea is that if something completes a task (like filling a tank) in 'T' units of time, its rate of work is 1/T per unit of time.

• If multiple agents (like pumps) work together to fill, their filling rates are added.
• If some agents (like leaks) work to empty while others fill, the net rate is the sum of filling rates minus the sum of emptying rates.
• If the net rate is positive, the tank fills. If it's negative, the tank empties.
• The time taken to complete the task (filling or emptying the full tank) is 1 divided by the net rate.

Understanding rates helps simplify these types of problems by converting the time taken into work done per unit of time.