Two pipes A and B can separately fill a cistern in 60 minutes and 75 minutes respectively. A third pipe at the bottom of the cistern can empty it. If all three pipes are opened together and the cistern becomes full in 50 minutes, in how much time can the third pipe alone empty the full cistern?

Introduction / Context:
This question is a classic pipes and cistern problem in quantitative aptitude. It tests the student's understanding of how to add and subtract rates of filling and emptying when multiple pipes work together. The key idea is that filling pipes contribute positive volume per minute, while an emptying pipe reduces the net rate. Interpreting the situation in terms of "work done per minute" makes the algebra straightforward and also matches many other work and time questions seen in exams.

Given Data / Assumptions:

Two filling pipes A and B fill the cistern in 60 minutes and 75 minutes respectively.
A third pipe C at the bottom empties the cistern at some constant rate.
All three pipes running together fill the cistern completely in 50 minutes from empty.
Flow rates of all pipes are constant over time and the tank capacity is fixed and non leaking except for pipe C.

Concept / Approach:
The standard approach is to treat one full cistern as one unit of work. A pipe that fills the tank in T minutes has a rate of 1/T tank per minute. When pipes work together, their rates are added if they are all filling. If one pipe is emptying, its rate is subtracted from the sum of the filling rates. We then equate the net rate to the observed combined time and solve for the unknown emptying time of pipe C.

Step-by-Step Solution:
Let the capacity of the cistern be 1 unit.Rate of pipe A = 1/60 unit per minute.Rate of pipe B = 1/75 unit per minute.Let the time taken by the emptying pipe C alone to empty the full cistern be x minutes.Then the rate of pipe C = 1/x unit per minute (emptying, so it will be subtracted).When all three pipes run together, the net rate is 1/60 + 1/75 - 1/x unit per minute.The cistern is filled in 50 minutes, so the net rate is also 1/50 unit per minute.Set up the equation: 1/60 + 1/75 - 1/x = 1/50.Compute 1/60 + 1/75 = (5 + 4) / 300 = 9/300 = 3/100.So 3/100 - 1/x = 1/50.Subtract 1/50 from both sides: 3/100 - 1/50 = 1/x.Write 1/50 as 2/100, so 3/100 - 2/100 = 1/100.Thus 1/x = 1/100, so x = 100 minutes.
Verification / Alternative check:
Check the net rate using x = 100 minutes.Rate of C = 1/100 unit per minute.Net rate = 1/60 + 1/75 - 1/100.Convert to a common denominator of 300: 1/60 = 5/300, 1/75 = 4/300, 1/100 = 3/300.Net rate = (5 + 4 - 3) / 300 = 6/300 = 1/50 unit per minute.At this rate, the time to fill the cistern is 50 minutes, which matches the problem statement. So the value x = 100 minutes is confirmed.
Why Other Options Are Wrong:
Options 80 minutes and 90 minutes would give a faster emptying pipe, which would reduce the net filling rate too much and lead to a filling time longer than 50 minutes.
Option 120 minutes makes the emptying pipe too slow, increasing the net rate and giving a filling time less than 50 minutes.
Option 150 minutes would slow the emptying effect even more and is inconsistent with the given combined time of 50 minutes.

Common Pitfalls:
Many learners mistakenly add the emptying pipe rate instead of subtracting it, which would produce an incorrect equation. Another frequent error is to confuse minutes and hours or to treat 50 minutes as 1/50 of a minute. Some students also try to manipulate fractions too quickly and make arithmetic errors when finding common denominators. Writing each step clearly with a common denominator avoids these mistakes. It is also important not to confuse the time taken by the combined system with the time taken by a single pipe.

Final Answer:
The third pipe alone can empty the full cistern in 100 minutes.