Understanding the Pipe, Tank, and Leak Scenario

This problem involves pipes filling a tank and a leak draining it. We need to determine how long it would take the leak alone to empty a full tank. The scenario happens in two phases: first, pipes A and B fill the tank while the leak is active (unnoticed), and second, the pipes continue filling after the leak is sealed.

Calculating Filling Rates of Pipes

First, let's find the rate at which each pipe fills the tank:

• Pipe A fills the tank in 36 minutes. So, in one minute, Pipe A fills \(\frac{1}{36}\) of the tank.
• Pipe B fills the tank in 45 minutes. So, in one minute, Pipe B fills \(\frac{1}{45}\) of the tank.

When both pipes A and B are opened simultaneously, their combined filling rate is the sum of their individual rates:

Combined rate of A and B \( = \frac{1}{36} + \frac{1}{45} \)

To add these fractions, we find a common denominator, which is the Least Common Multiple (LCM) of 36 and 45. The LCM of 36 and 45 is 180.

\( \frac{1}{36} + \frac{1}{45} = \frac{1 \times 5}{36 \times 5} + \frac{1 \times 4}{45 \times 4} = \frac{5}{180} + \frac{4}{180} = \frac{5+4}{180} = \frac{9}{180} = \frac{1}{20} \)

So, pipes A and B together can fill \(\frac{1}{20}\) of the tank in one minute.

Analyzing the Tank Filling in Two Phases

Let \(L\) be the time in minutes it takes the leak alone to empty the full tank. The rate of the leak is then \(\frac{1}{L}\) of the tank per minute (this is a draining rate, so it's negative work).

Phase 1: First 20 minutes (Pipes A, B, and Leak active)

In the first 20 minutes, both pipes A and B were filling, and the leak was draining. The net rate of work (filling - draining) during this phase was:

Net rate \( = \) (Rate of A + Rate of B) \( - \) (Rate of Leak) \( = \frac{1}{20} - \frac{1}{L} \)

The portion of the tank filled in these 20 minutes is:

Work done in Phase 1 \( = \text{Net rate} \times \text{Time} = \left(\frac{1}{20} - \frac{1}{L}\right) \times 20 = 20 \times \frac{1}{20} - 20 \times \frac{1}{L} = 1 - \frac{20}{L} \)

Phase 2: Next 15 minutes (Pipes A and B active, Leak sealed)

After 20 minutes, the leak was sealed. Only pipes A and B were working for the next 15 minutes at their combined rate of \(\frac{1}{20}\) tank per minute.

The portion of the tank filled in these 15 minutes is:

Work done in Phase 2 \( = \) (Combined rate of A and B) \( \times \) Time \( = \frac{1}{20} \times 15 = \frac{15}{20} = \frac{3}{4} \)

Total Work to Fill the Tank

The tank was completely filled after these two phases. This means the sum of the work done in Phase 1 and Phase 2 equals the filling of one full tank (which is represented by 1).

Total Work \( = \) Work done in Phase 1 \( + \) Work done in Phase 2 \( = 1 \)

\( \left(1 - \frac{20}{L}\right) + \frac{3}{4} = 1 \)

Solving for the Leak Emptying Time (L)

Now we solve the equation for \(L\):

\( 1 - \frac{20}{L} + \frac{3}{4} = 1 \)

Subtract 1 from both sides of the equation:

\( -\frac{20}{L} + \frac{3}{4} = 0 \)

Add \(\frac{20}{L}\) to both sides:

\( \frac{3}{4} = \frac{20}{L} \)

Cross-multiply:

\( 3 \times L = 4 \times 20 \)

\( 3L = 80 \)

Divide by 3:

\( L = \frac{80}{3} \)

To express this as a mixed number, divide 80 by 3:

\( 80 \div 3 = 26 \) with a remainder of \( 2 \).

So, \( \frac{80}{3} = 26\frac{2}{3} \).

Final Answer for Leak Emptying Time

The time taken for the leak alone to empty the full tank is \(26\frac{2}{3}\) minutes.

Revision Table: Key Concepts in Tank and Pipe Problems

Additional Information: Solving Time and Work Problems

Problems involving pipes and tanks are similar to time and work problems. Here are some key ideas:

• Individual entities (pipes, taps, people) have a rate of work.
• Rates are usually expressed as "amount of work per unit of time" (e.g., \(\frac{1}{36}\) of tank per minute).
• Filling pipes do positive work, while draining pipes or leaks do negative work.
• When multiple entities work together, their rates are added (or subtracted for negative work).
• The total work done to complete a task (like filling a tank) is typically represented as 1 unit.
• Break down complex problems into phases if the working entities or their rates change over time. Calculate the work done in each phase and sum them up to equal the total work (1).