Pipes A and B can fill a tank in 60 minutes and 90 minutes respectively. There is a leak at three-fourths of the height of the tank. If the leak alone is opened when the tank is full, it takes 36 minutes to lower the water level from full down to three-fourths of the height. If both pipes and the leak are opened simultaneously from an empty tank, how many minutes will it take to fill the tank completely?

Introduction / Context:
This problem combines multiple ideas: two inlet pipes, one leak, and the fact that the leak only operates when the water level is above three-fourths of the tank height. We must carefully separate the stages of filling and use the information about how fast the leak can lower the water level from full to three-fourths of the height.

Given Data / Assumptions:

Pipe A fills the full tank in 60 minutes.

Pipe B fills the full tank in 90 minutes.

The leak is located at three-fourths of the tank height.

When the tank is full and only the leak is open, it takes 36 minutes to lower the level to three-fourths of the height.

The tank has uniform cross section, so volume is proportional to height.

Concept / Approach:
Because the tank cross section is uniform, lowering the level from full to three-fourths of the height corresponds to removing one-fourth of the tank volume. Thus the leak alone empties one-fourth of the tank in 36 minutes. With two inlets and one leak, we consider two stages: from empty to three-fourths of the tank (when the leak is not yet active), and from three-fourths to full (when the leak is active along with the inlets).

Step-by-Step Solution:
Step 1: Let the full tank volume be 1 unit.Step 2: Rate of pipe A = 1 / 60 tank per minute.Step 3: Rate of pipe B = 1 / 90 tank per minute.Step 4: Leak alone removes 1 / 4 of the tank in 36 minutes, so leak rate = (1 / 4) / 36 = 1 / 144 tank per minute.Step 5: Combined filling rate of A and B = 1 / 60 + 1 / 90 = (3 + 2) / 180 = 5 / 180 = 1 / 36 tank per minute.Step 6: First stage: from empty to three-fourths of the tank. In this stage the leak does not operate, so the rate is 1 / 36.Step 7: Time for stage one = (3 / 4) / (1 / 36) = (3 / 4) * 36 = 27 minutes.Step 8: Second stage: from three-fourths to full. Now both inlets and the leak are active.Step 9: Net rate in stage two = (1 / 36) (A and B) minus 1 / 144 (leak) = (4 / 144 - 1 / 144) = 3 / 144 = 1 / 48 tank per minute.Step 10: Volume to fill in stage two = 1 - 3 / 4 = 1 / 4 tank.Step 11: Time for stage two = (1 / 4) / (1 / 48) = (1 / 4) * 48 = 12 minutes.Step 12: Total time to fill the tank = 27 + 12 = 39 minutes.

Verification / Alternative check:
We can confirm the leak rate: if it empties at 1 / 144 tank per minute, in 36 minutes it empties 36 / 144 = 1 / 4 of the tank, which matches the given drop from full to three-fourths level. Using this rate in the net filling computation gives a reasonable total time less than one hour, consistent with the speeds of the pipes.

Why Other Options Are Wrong:
27 minutes corresponds only to the first stage and ignores the slower second stage where the leak counteracts the inlets.

33 minutes and 37 minutes both underestimate the contribution of the leak.

Any value smaller than 39 minutes would imply a faster net filling rate during the second stage than is actually possible given the leak rate.

Common Pitfalls:
One common error is to assume the leak operates from the beginning, rather than only after the water reaches its height. Another is to misinterpret “three-fourths of the height” as something other than three-fourths of the volume in a uniform tank. Also, some students forget to compute the net rate correctly when the leak is active. Properly dividing the process into two phases avoids these mistakes.

Final Answer:
The tank will be completely filled in 39 minutes.