Solving the Tank Filling Problem with Leakage

This problem involves calculating the time taken to fill a tank when there's both an inflow (filling pipe) and an outflow (leakage). We need to determine the time it takes for the leakage alone to empty a specific portion of the tank.

Understanding Filling and Emptying Rates

Let's denote the total capacity of the tank as $V$ units.

• The rate at which the filling pipe works is the volume it fills per hour.
• The rate at which the leak works is the volume it empties per hour.

Step 1: Calculate the Filling Pipe's Rate

The filling pipe normally fills the tank in 18 hours. So, its filling rate ($R_f$) is:

Rate of filling pipe, $R_f = \frac{\text{Volume}}{\text{Time}} = \frac{V}{18}$ units per hour.

Step 2: Determine the Effective Filling Time

Due to a leakage, the tank takes 6 hours *more* to fill. This means the total time taken to fill the tank with both the pipe filling and the leak emptying is:

Effective filling time = 18 hours + 6 hours = 24 hours.

Step 3: Calculate the Combined Filling and Leakage Rate

In these 24 hours, the tank is filled completely ($V$ units). The combined rate of the filling pipe and the leak working together is:

Combined Rate = $\frac{\text{Volume}}{\text{Effective Time}} = \frac{V}{24}$ units per hour.

This combined rate is the difference between the filling pipe's rate and the leak's rate (since the leak reduces the net filling). Let the leak's emptying rate be $R_l$. So,

$R_f - R_l = \frac{V}{24}$

Step 4: Calculate the Leakage Rate ($R_l$)

We know $R_f = \frac{V}{18}$. Substitute this into the equation from Step 3:

$$ \frac{V}{18} - R_l = \frac{V}{24} $$

Now, solve for $R_l$:

$$ R_l = \frac{V}{18} - \frac{V}{24} $$

To subtract these fractions, find a common denominator, which is 72:

$$ R_l = V \left( \frac{1}{18} - \frac{1}{24} \right) $$

$$ R_l = V \left( \frac{4}{72} - \frac{3}{72} \right) $$

$$ R_l = V \left( \frac{1}{72} \right) = \frac{V}{72} $$

This means the leakage alone can empty the entire tank ($V$ units) in 72 hours.

Step 5: Calculate Time to Empty Two-Thirds of the Tank

The question asks for the time the leakage takes to empty two-thirds of the tank. The volume to be emptied is $\frac{2}{3}V$. Using the leak's rate ($R_l = \frac{V}{72}$):

$$ \text{Time to empty } \frac{2}{3}V = \frac{\text{Volume to empty}}{\text{Leakage Rate}} $$

$$ \text{Time} = \frac{\frac{2}{3}V}{\frac{V}{72}} $$

$$ \text{Time} = \frac{2V}{3} \times \frac{72}{V} $$

The $V$ terms cancel out:

$$ \text{Time} = \frac{2}{3} \times 72 $$

$$ \text{Time} = 2 \times \frac{72}{3} $$

$$ \text{Time} = 2 \times 24 $$

$$ \text{Time} = 48 \text{ hours} $$

Conclusion

The leakage point can empty two-thirds of the tank in 48 hours.