Pipes and Cistern problems generally consist of a cistern (tank) to which one or more pipes fill the cistern or empty the cistern. These problems of pipes and cisterns can be solved by using the same method used in time and work. And we changes out formulae according to the requirement of the pipes and cisterns.

**Outlet:** The pipe connected to the cistern, which helps to empty the tank Â is called an outlet.

**Inlet:** The pipe which is connected withÂ the tank, that fills it is called an inlet.

Pipes and Cisterns are somewhat similar to the concepts of Work and Wages:

- The problems of pipes and cisterns usually have two kinds of pipes, Inlet pipe and Outlet pipe / Leak. Inlet pipe is the pipe that fills the tank / reservoir / cistern and Outlet pipe / Leak is the one that empties it.
- If a pipe can fill a tank in â€˜nâ€™ hours, then in 1 hour, it will fill ‘\(\frac{1}{n}\)‘ parts. For example, if a pipe takes 6 hours to fill a tank completely, say of 12 liters, then in 1 hour, it will fill \((\frac{1}{6})^{th}\) of the tank, i.e., 2 liters.
- If a pipe can empty a tank in â€˜nâ€™ hours, then in 1 hour, it will empty ‘\(\frac{1}{n}\)‘ parts. For example, if a pipe takes 6 hours to empty a tank completely, say of 18 liters, then in 1 hour, it will empty \((\frac{1}{6})^{th}\) of the tank, i.e., 3 liters.
- If we have a number of pipes such that some fill the tank and some empty it, and we open all of them together, then in one hour, part of the tank filled/emptied = \(\sum(\frac{1}{m_{i}})\) – \(\sum(\frac{1}{n_{j}})\) where ‘\(m_{i}\)‘ is the time taken by inlet pipe ‘i’ to fill the tank completely if only it were open and ‘\(n_{j}\)‘ is the time taken by outlet pipe â€˜jâ€™ to empty the tank completely if only it were open. If the sign of this equation is positive, the tank would be filled and if the sign is negative, the tank would be emptied.

**Example 1**:

Two pipes A and B can fill a tank separately in 12 and 16 hours respectively. If both of them are opened together when the tank is initially empty, how much time will it take to completely fill the tank?

**Solution**:

**Method 1**:

- Part of tank filled by pipe A in one hour working alone = \(\frac{1}{12}\)

Part of tank filled by pipe B in one hour working alone = \(\frac{1}{16}\)

⇒ Part of tank filled by pipe A and pipe B in one hour working together = (\(\frac{1}{12}\)) + (\(\frac{1}{16}\)) = \(\frac{7}{48}\)

Therefore, time taken to completely fill the tank if both A and B work together = \(\frac{48}{7}\) hours

**Method 2**:

- Let the capacity of tank be LCM (12, 16) = 48 units

⇒ Efficiency of pipe A = \(\frac{48}{12}\) = 4 units / hour

⇒ Efficiency of pipe B = \(\frac{48}{16}\) = 3 units / hour

⇒ Combined efficiency of pipes A and B = 7 units / hour

Therefore, time taken to completely fill the tank = \(\frac{48}{7}\) hours

**Example 2**:

Three pipes A, B and C are connected to a tank. Out of the three, A is the inlet pipe and B and C are the outlet pipes. If opened separately, A fills the tank in 10 hours, B empties the tank in 12 hours and C empties the tank in 30 hours. If all three are opened simultaneously, how much time does it take to fill / empty the tank?

**Solution**:

**Method 1**:

- Part of tank filled by pipe A in one hour working alone = \(\frac{1}{10}\)

Part of tank emptied by pipe B in one hour working alone = \(\frac{1}{12}\)

Part of tank emptied by pipe B in one hour working alone = \(\frac{1}{30}\)

⇒ Part of tank filled by pipes A, B and C in one hour working together = (\(\frac{1}{10}\)) â€“ (\(\frac{1}{12}\)) â€“ (\(\frac{1}{30}\)) = –\(\frac{1}{60}\)

Therefore, time taken to completely empty the tank if all pipes are opened simultaneously = \(\frac{1}{60}\) hours = 60 hours

**Method 2**:

- Let the capacity of tank be LCM (10, 12, 30) = 60 units

⇒ Efficiency of pipe A = \(\frac{60}{10}\) = 6 units / hour

⇒ Efficiency of pipe B = – \(\frac{60}{10}\) = – 5 units / hour (Here, â€˜-â€˜ represents outlet pipe)

⇒ Efficiency of pipe C = – \(\frac{60}{30}\) = – 2 units / hour (Here, â€˜-â€˜ represents outlet pipe)

⇒ Combined efficiency of pipes A, B and C = 6 â€“ 5 â€“ 2 = â€“ 1 units / hour (Here, â€˜-â€˜represents outlet pipe)

Therefore, time taken to completely empty the tank = \(\frac{60}{(1)}\) = 60 hours

**Example 3**:

A cistern has two pipes. Both working together can fill the cistern in 12 minutes. First pipe is 10 minutes faster than the second pipe. How much time would it take to fill the cistern if only second pipe is used?

**Solution**:

**Method 1**:

- Let the time taken by first pipe working alone be â€˜tâ€™ minutes.

⇒ Time taken by second pipe working alone = t + 10 minutes.

Part of tank filled by pipe A in one hour working alone = \(\frac{1}{t}\)

Part of tank filled by pipe B in one hour working alone = \(\frac{1}{t + 10}\)

=> Part of tank filled by pipe A and B in one hour working together = \(\frac{1}{t}\) + \(\frac{1}{t + 10}\) = \(\frac{2t + 10}{t \times (t + 10)}\)

But we are given that it takes 12 minutes to completely fill the cistern if both pipes are working together.

⇒ \(\frac{2t + 10}{t \times (t + 10)}\) = \(\frac{1}{12}\)

⇒ \(\frac{t \times (t + 10)}{2t + 10}\) = 12

⇒ \(t^{2}\) + 10t = 24t +120

⇒ \(t^{2}\) â€“ 14t – 120 = 0

⇒ (t -20)(t + 6) = 0

⇒ t = 20 minutes (Time cannot be negative)

Therefore, time taken by second pipe working alone = 20 + 10 = 30 minutes

**Method 2**:

- Let the time taken by first pipe working alone be â€˜tâ€™ minutes.

=> Time taken by second pipe working alone = t + 10 minutes.

Let the capacity of cistern be t x (t + 10) units.

=> Efficiency of first pipe = \(\frac{t \times (t + 10)}{t}\) = (t + 10) units / minute

=> Efficiency of second pipe = \(\frac{t \times (t + 10)}{(t + 10)}\) = t units / minute

=> Combined efficiency of pipes = (2t + 10) units / minute

=> Time taken to fill the cistern completely = \(\frac{t \times (t + 10)}{(2t + 10)}\)

But we are given that it takes 12 minutes to completely fill the cistern if both pipes are working together.

\(\frac{t \times (t + 10)}{(2t + 10)}\) = 12

\(t^{2}\) + 10t = 24t +120

\(t^{2}\) – 14t – 120 = 0

(t â€“ 20)(t + 6) = 0

t = 20 minutes (Time cannot be negative)

Therefore, time taken by second pipe working alone = 20 + 10 = 30 minutes

part filled in 1 hour = \(\frac{1}{x}\)

**2.** If the pipe can empty a full tank inÂ \(y\) hours, then:

part emptied in 1 hour = \(\frac{1}{y}\)

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

**4.** If a pipe can fill a tank in \(x\) hours and another pipe can empty the full tank in \(y\) (where \(x\) > \(y\)), then on opening both the pipes, the net part emptiedÂ in 1 hour = (\(\frac{1}{y}\) – \(\frac{1}{x}\))

**Solution**:

- Given,

A pipe fills a tank in 36 hours

â‡’ part filled by A in 1 hour = \(\frac{1}{36}\)

B pipe fills a tank in 45 hours

â‡’ part filled by B in 1 hour = \(\frac{1}{45}\)

Part filled by both in 1 hour = \(\frac{1}{36}\) + \(\frac{1}{45}\)

â‡’ \(\frac{9}{180}\)

â‡’ \(\frac{1}{20}\)

Therefore, both the pipes fills the tank together in 20 hours

**2. Two pipes can fill a tank in 12 hours and 14 hours respectively while a third pipe empties the full tank in 18 hours. If all the three pipes operate simultaneously, in how much time will the tank be filled?**

**Solution**:

- Given that,

First pipe fills the tank in 12 hours.

Second pipe fills the tank in 14 hours.

Third pipe empties it in 18 hours.

Total part filled in 1 hour = \(\frac{1}{12}\) + \(\frac{1}{14}\) – \(\frac{1}{18}\)

â‡’ \(\frac{21 + 18 – 14}{252}\)

â‡’ \(\frac{25}{252}\)

Therefore, the tank filled in 10.08 hours

**3. A pump can fill a tank in 2 hours. Due to leakage, it took 2\(\frac{2}{3}\) hours to fill the tank. How much time will leak can drain all the water of tank in?**

**Solution**:

- Let the leak drain all the water of tank = \(y\) hours

Part filled by the pipe in one hour = \(\frac{1}{2}\)

Part emptied by the leak in one hour = \(\frac{1}{y}\)

Now, \(\frac{1}{2}\) – \(\frac{1}{y}\) = \(\frac{3}{8}\)

â‡’ \(\frac{1}{y}\) = \(\frac{1}{2}\) – \(\frac{3}{8}\)

â‡’ \(\frac{1}{y}\) = \(\frac{4 -3}{8}\)

â‡’ \(\frac{1}{y}\) = \(\frac{1}{8}\)

â‡’ \(y\) = 8

Therefore, the leak took 8 hours to drain the full tank.

**4. 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.**

**Solution**:

- Part of tank filled by A in 1 hour = \(\frac{1}{30}\)

Part of tank filled by B in 1 hour = \(\frac{1}{20}\)

Part of tank filled by C in 1 hour = \(\frac{1}{10}\)

Pipe C discharges chemical solution R. So,

Part of tank filled by C in 3 minutes = 3 x \(\frac{1}{10}\) = \(\frac{3}{10}\)

Part filled by all the three in 1 hour = \(\frac{1}{30}\) + \(\frac{1}{20}\) + \(\frac{1}{10}\)

â‡’ \(\frac{2 + 3 + 6}{60}\)

â‡’ \(\frac{11}{60}\)

Part filled by all the three in 3 minutes = 3 x \(\frac{11}{60}\)

â‡’ \(\frac{11}{20}\)

Hence, \(\frac{(\frac{3}{10})}{(\frac{11}{20})}\) = \(\frac{(3 * 20)}{(11 * 10)}\) = \(\frac{6}{11}\)

Therefore, the proportion of solution R in the liquid after 3 minutes = \(\frac{6}{11}\)

**5. A cistern can be filled by a tap in 3 hours while it can be emptied by another tap in 8 hours. If both the taps are opened simultaneously, then by how much time will the cistern gets filled?**

**Solution**:

- Given that,

First tap fills in 3 hours = Part filled by the first tap in 1 hour = \(\frac{1}{3}\)

Second tap fills in 8 hours = Part filled by second tap in 1 hour

= \(\frac{1}{8}\)

Total filled by both in 1 hour = \(\frac{1}{3}\) + \(\frac{1}{8}\) = \(\frac{5}{24}\)

Therefore, the cistern gets filled in \(\frac{24}{5}\) = 4.8 hours