Solving Water Tank Filling Time with Taps and Leakage

This problem involves calculating the time taken to fill a water tank when multiple inputs (taps) and outputs (leakage) are operating simultaneously, with a change occurring during the process.

The core concept here is understanding the 'rate' of filling or emptying. The rate is the amount of work done (filling or emptying a tank) per unit of time. If a tap fills a tank in \(T\) hours, its filling rate is \( \frac{1}{T} \) tank per hour. Similarly, if a leakage empties a tank in \(E\) hours, its emptying rate is \( \frac{1}{E} \) tank per hour (considered negative for combined work).

Calculating Individual Rates

• Tap 1 fills the tank in 12 hours. Its filling rate is \( \frac{1}{12} \) tank/hour.
• Tap 2 fills the tank in 18 hours. Its filling rate is \( \frac{1}{18} \) tank/hour.
• The leakage empties the tank in 36 hours. Its emptying rate is \( -\frac{1}{36} \) tank/hour.

Analyzing the First Hour

For the first hour, both water taps are open, and the leakage point is also active. So, the net rate of filling the tank during this hour is the sum of the rates of Tap 1 and Tap 2 minus the rate of the leakage.

Net rate in the first hour \( = \) Rate of Tap 1 \( + \) Rate of Tap 2 \( + \) Rate of Leakage

\( = \frac{1}{12} + \frac{1}{18} - \frac{1}{36} \)

To add and subtract these fractions, we find a common denominator, which is 36.

\( = \frac{1 \times 3}{12 \times 3} + \frac{1 \times 2}{18 \times 2} - \frac{1 \times 1}{36 \times 1} \)

\( = \frac{3}{36} + \frac{2}{36} - \frac{1}{36} \)

\( = \frac{3 + 2 - 1}{36} \)

\( = \frac{4}{36} \)

\( = \frac{1}{9} \) tank/hour.

So, in the first hour, the amount of tank filled is:

Amount filled in 1 hour \( = \) Net rate \( \times \) Time

\( = \frac{1}{9} \times 1 = \frac{1}{9} \) of the tank.

Filling the Remaining Tank After Leakage Repair

After 1 hour, the leakage point is repaired. This means that only the two water taps are working to fill the rest of the tank.

The remaining part of the tank to be filled is \( 1 - \frac{1}{9} \).

Remaining part \( = \frac{9}{9} - \frac{1}{9} = \frac{8}{9} \) of the tank.

The combined filling rate of the two taps is:

Combined rate of two taps \( = \) Rate of Tap 1 \( + \) Rate of Tap 2

\( = \frac{1}{12} + \frac{1}{18} \)

Finding a common denominator (36):

\( = \frac{1 \times 3}{12 \times 3} + \frac{1 \times 2}{18 \times 2} \)

\( = \frac{3}{36} + \frac{2}{36} \)

\( = \frac{3 + 2}{36} \)

\( = \frac{5}{36} \) tank/hour.

Now, we calculate the time taken to fill the remaining \( \frac{8}{9} \) part of the tank at this combined rate of \( \frac{5}{36} \) tank/hour.

Time taken for remaining part \( = \frac{\text{Remaining part}}{\text{Combined rate of two taps}} \)

\( = \frac{8/9}{5/36} \)

\( = \frac{8}{9} \times \frac{36}{5} \)

Cancel out 9 from 36:

\( = \frac{8}{1} \times \frac{4}{5} \)

\( = \frac{32}{5} \) hours.

Calculating Total Time

The total time to fill the tank is the time taken in the first hour plus the time taken to fill the remaining part.

Total time \( = \) Time in 1st hour \( + \) Time for remaining part

\( = 1 \) hour \( + \frac{32}{5} \) hours.

Let's convert \( \frac{32}{5} \) hours into hours and minutes.

\( \frac{32}{5} = 6 \frac{2}{5} \) hours.

So, this is 6 hours and \( \frac{2}{5} \) of an hour. To convert \( \frac{2}{5} \) hours to minutes, multiply by 60:

\( \frac{2}{5} \times 60 \) minutes \( = 2 \times 12 \) minutes \( = 24 \) minutes.

Thus, the time taken for the remaining part is 6 hours and 24 minutes.

Total time \( = 1 \) hour \( + 6 \) hours 24 minutes \( = 7 \) hours 24 minutes.

Summary of Tank Filling Calculation

Here is a breakdown of the steps and calculations:

The empty tank will be completely filled in a total of 7 hours and 24 minutes.

Revision Table: Water Tank Calculations

Additional Information: Time and Work Problems

Problems involving filling tanks with taps and leakages are common examples of 'Time and Work' problems. Key principles include:

• The total work is usually considered as '1 unit' (e.g., filling one tank).
• The rate of work is the amount of work done in one unit of time.
• If multiple agents work together, their rates are added (or subtracted if they work against each other, like a leakage).
• If the working condition changes (like repairing the leakage), the problem needs to be solved in phases. Calculate the work done in the first phase and the time taken. Then calculate the remaining work and the rate for the second phase to find the time for the remaining work. The total time is the sum of the times for all phases.
• Understanding fractions is crucial for calculating parts of the work done and remaining work.

These principles apply to various scenarios, such as multiple pipes filling or emptying a tank, or multiple people completing a task at different speeds.