Pipes and Cisterns: Complete Master Guide for UPSC/MPSC CSAT (English)

Introduction to Pipes and Cisterns

The concept of “Pipes and Cisterns” is an integral part of the Quantitative Aptitude syllabus for competitive exams like UPSC CSAT and MPSC. At its core, the logic is identical to that of “Time and Work.” In Time and Work, we calculate the work done by individuals, whereas in Pipes and Cisterns, we measure the amount of water filled or emptied by pipes in a tank or reservoir (cistern).

Understanding this topic is crucial because it frequently appears in examinations and tests your logical deduction, fraction calculation, and ability to comprehend varying rates of work. The primary difference from Time and Work is the introduction of “negative work,” representing pipes that empty the tank.

Core Concepts and Formulas

Before diving into problems, it is important to understand a few fundamental terms and the rules governing them:

• Inlet: A pipe connected with a tank or cistern that fills it is known as an inlet. The work done by an inlet is always considered positive.
• Outlet: A pipe connected with a tank or cistern that empties it is known as an outlet. The work done by an outlet is always considered negative.

Important Formulas

• If a pipe can fill a tank in $x$ hours, then the part of the tank filled in 1 hour = $\frac{1}{x}$.
• If a pipe can empty a full tank in $y$ hours, then the part of the tank emptied in 1 hour = $\frac{1}{y}$.
• If a pipe can fill a tank in $x$ hours and another pipe can empty the full tank in $y$ hours (where $y > x$), then on opening both the pipes, the net part filled in 1 hour = $\left(\frac{1}{x} – \frac{1}{y}\right)$.
• If a pipe can fill a tank in $x$ hours and another pipe can empty the full tank in $y$ hours (where $x > y$), then on opening both the pipes, the net part emptied in 1 hour = $\left(\frac{1}{y} – \frac{1}{x}\right)$.

Pro-Tips to Avoid Common Mistakes

• Use the LCM Method: Instead of dealing with fractions (like $\frac{1}{x}$), assume the total capacity of the tank to be the Least Common Multiple (LCM) of the given times. Calculate the efficiency (units of water per hour/minute) of each pipe. This significantly speeds up calculation.
• Mind the Signs: Always double-check if a pipe is an inlet (+) or an outlet (-). A common error is adding an outlet’s efficiency instead of subtracting it.
• Read Carefully: Pay attention to whether the question asks for the time to fill the remaining tank or the total time to fill the tank.
• Initial State of the Tank: Note if the tank is initially empty, partially full, or completely full.

Solved Examples with Step-by-Step Explanations

Example 1: Basic Inlet Pipes

Question: Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. If both pipes are opened together, how much time will they take to fill the tank completely?

Solution:

• Method 1 (Fraction): Part filled by A in 1 min = $\frac{1}{20}$. Part filled by B in 1 min = $\frac{1}{30}$. Part filled by (A + B) in 1 min = $\frac{1}{20} + \frac{1}{30} = \frac{3+2}{60} = \frac{5}{60} = \frac{1}{12}$. Therefore, both pipes together can fill the tank in 12 minutes.
• Method 2 (LCM Approach): Let the total capacity of the tank be the LCM of 20 and 30 = 60 units. Efficiency of A = $\frac{60}{20} = 3$ units/min. Efficiency of B = $\frac{60}{30} = 2$ units/min. Total efficiency (A + B) = $3 + 2 = 5$ units/min. Total time required = $\frac{\text{Total Capacity}}{\text{Total Efficiency}} = \frac{60}{5} = 12$ minutes.

Example 2: Inlet and Outlet Working Together

Question: Pipe A can fill a tank in 10 hours. Pipe B can empty the same tank in 15 hours. If both pipes are opened simultaneously, how much time will it take to fill the empty tank?

Solution:

• Total capacity of the tank = LCM of 10 and 15 = 30 units.
• Efficiency of A (Inlet) = $\frac{30}{10} = +3$ units/hour.
• Efficiency of B (Outlet) = $\frac{30}{15} = -2$ units/hour.
• Effective efficiency (A + B) = $3 + (-2) = +1$ unit/hour (Since it’s positive, the tank is being filled).
• Time required to fill the tank = $\frac{30}{1} = 30$ hours.

Example 3: Leakage Problem

Question: A pump can fill a tank with water in 2 hours. Because of a leak in the tank, it took $2\frac{1}{3}$ hours to fill the tank. If the tank is full, how long will it take for the leak to empty it?

Solution:

• Time taken by Pump (P) without leak = 2 hours.
• Time taken by Pump (P) + Leak (L) = $2\frac{1}{3} = \frac{7}{3}$ hours.
• Let’s use fractions here. Part filled by P in 1 hour = $\frac{1}{2}$.
• Part filled by (P + L) in 1 hour = $\frac{1}{7/3} = \frac{3}{7}$.
• Work done by Leak in 1 hour = (Part filled by P+L) – (Part filled by P) = $\frac{3}{7} – \frac{1}{2} = \frac{6 – 7}{14} = -\frac{1}{14}$.
• The negative sign indicates the leak is emptying the tank. It empties $\frac{1}{14}$ of the tank in 1 hour.
• Therefore, the leak will empty the full tank in 14 hours.

Example 4: Pipe Closed After Sometime

Question: Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both pipes are opened together, but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

Solution:

• Total capacity = LCM of 15 and 20 = 60 units.
• Efficiency of A = $\frac{60}{15} = 4$ units/min.
• Efficiency of B = $\frac{60}{20} = 3$ units/min.
• For the first 4 minutes, both A and B are open. Work done in 4 mins = $4 \times (4 + 3) = 4 \times 7 = 28$ units.
• Remaining capacity to be filled = $60 – 28 = 32$ units.
• Now, pipe A is closed. The remaining 32 units will be filled by pipe B alone.
• Time taken by B to fill remaining part = $\frac{32}{3} = 10\frac{2}{3}$ minutes.
• Total time taken = Initial 4 mins + $10\frac{2}{3}$ mins = $14\frac{2}{3}$ minutes.

Example 5: Multiple Similar Taps

Question: A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?

Solution:

• Time taken by one tap to fill the whole tank = 6 hours.
• Time taken by one tap to fill half the tank = 3 hours.
• Remaining part of the tank = $\frac{1}{2}$.
• Now, 3 more similar taps are opened, making a total of 4 taps.
• Since 1 tap takes 6 hours to fill the full tank, 4 taps will take $\frac{6}{4} = \frac{3}{2}$ hours = 1 hour 30 minutes to fill the full tank.
• Therefore, 4 taps will take half of this time to fill the remaining half tank = $\frac{1.5}{2}$ hours = $0.75$ hours = $45$ minutes.
• Alternatively: Part filled by 4 taps in 1 hour = $4 \times \left(\frac{1}{6}\right) = \frac{2}{3}$. Time required to fill $\frac{1}{2}$ part = $\frac{1/2}{2/3} = \frac{3}{4}$ hours = 45 minutes.
• Total time taken = 3 hours (initial) + 45 minutes = 3 hours 45 minutes.

Practice Questions

1. Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P, Q and R respectively. What is the proportion of solution R in the liquid in the tank after 3 minutes?
2. Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 minutes?
3. A cistern is normally filled in 8 hours but takes two hours longer to fill because of a leak in its bottom. If the cistern is full, in how many hours will the leak empty it?
4. Pipe A can fill a tank in 16 minutes and pipe B empties it in 24 minutes. If both the pipes are opened together, after how many minutes should pipe B be closed, so that the tank is filled in 30 minutes?