Pipe and Cistern

Start with the simplest picture: one pipe, flowing at a constant rate. If it runs for double the time, it fills (or empties) double the water. So for a single pipe, time is directly proportional to the part of the tank.

That gives two moves:

• Part → time. If the full tank takes T minutes, then a part f of the tank takes f × T minutes. 13 of a 60-minute tank takes 20 minutes; 40% of it takes 24 minutes.
• Time → full tank. If a part f takes t minutes, then the full tank takes tf minutes. If 25 of the tank fills in 30 minutes, the full tank takes 30 ÷ 25 = 75 minutes.

[[fig:0]]

When the question gives one part and asks for another, either go through the full tank or use the ratio of the two parts directly.

Radhika's garden pipe fills 25 of a drum in 30 minutes. Full drum = 30 × 52 = 75 minutes. Half the drum = 12 × 75 = 3712 minutes. By the ratio of parts it is one step: 12 ÷ 25 = 54, so 30 × 54 = 3712 minutes.

[[fig:1]]

Emptying follows the same rule, because an outlet working alone is a pipe running in the other direction. The only real decision in this type is whether to multiply or divide.

When filling pipes run together, their efficiencies add. If A fills 16 of the tank per hour and B fills 112, together they fill 16 + 112 = 14 per hour, so the tank fills in 4 hours.

The capacity method makes it cleaner. Radhika's school tank has taps A (6 hours) and B (12 hours). Capacity = LCM(6, 12) = 12 units. A = 2 units/hour, B = 1 unit/hour, together 3 units/hour. Time = 123 = 4 hours.

[[fig:0]]

For exactly two pipes there is a well-known formula: time together = xyx + y. With 6 and 12 it gives 7218 = 4 hours. Keep it for two pipes; for three pipes it does not become xyzx+y+z. For three pipes add the reciprocals, 1T = 1x + 1y + 1z, or use LCM units. With 4, 6 and 12 hours: capacity 12, efficiencies 3 + 2 + 1 = 6, time = 2 hours.

Pairs given (A+B, B+C, C+A). Sometimes the times are given for pipes in pairs. Add all three pair-efficiencies: (A+B) + (B+C) + (C+A) = 2(A+B+C). Half of that sum is the efficiency of all three together. Then any single pipe = (all three) − (the pair without it). If A+B+C is given directly, it is quicker still: B = (A+B+C) − (C+A).

[[fig:1]]

Equation type. A few questions give no times at all, only relations such as "A takes 5 minutes more than B". You can form an equation in one unknown, but in the exam put the options in; the option that satisfies the condition is the answer.

This type has two faces, and both rest on the same fact: efficiency and time are inversely proportional.

Face 1: "one pipe is n times as fast as the other". If A is 4 times as fast as B, then in one minute A fills 4 units while B fills 1 unit. So the efficiencies are 4 : 1 and the times are 1 : 4. The question usually gives the time taken together. Together they fill 4 + 1 = 5 units per minute, so capacity = 5 × (time together). After that, any single pipe's time is capacity ÷ its units.

Radhika's two taps fill a bucket in 12 minutes together, and one tap is twice as fast as the other. Efficiencies 2 : 1, total 3 units/min, capacity = 3 × 12 = 36 units. Faster tap = 362 = 18 minutes, slower tap = 36 minutes.

[[fig:0]]

"One and a half times as fast" means 1.5 : 1 = 3 : 2. "Two and a half times as fast" means 2.5 : 1 = 5 : 2. Clear the decimal first. When the question says "A is twice as fast as B and B is thrice as fast as C", build the chain from the smallest: C = 1, B = 3, A = 6.

Face 2: many identical pipes. If all the pipes are of the same kind, count pipe-hours the way you count man-days. 6 pipes taking 10 hours means the tank needs 6 × 10 = 60 pipe-hours. With 4 such pipes it will take 604 = 15 hours. So

P₁ × T₁ = P₂ × T₂

[[fig:1]]

The same idea handles pipes that join in the middle: find the pipe-hours already used, and divide the rest by the new number of pipes. If the tank size changes, scale the pipe-hours by the size; if a pipe's efficiency becomes 13, it needs 3 times the time.

Remember one more rule. The flow through a pipe is proportional to the area of its cross-section, that is, to the square of its radius. A pipe of double radius delivers 4 times the water. Such a question is rare, but it is an easy mark if you know this.

[[fig:2]]

This is the new idea of the chapter. An outlet pipe works against the inlets, so its efficiency is written with a minus sign. With everything open together:

Net efficiency = (inlets) − (outlets)

Take Radhika's tank. Its filling pipe takes 9 hours, and a drain empties the full tank in 12 hours. Capacity = LCM(9, 12) = 36 units. Inlet = +4 units/hour, drain = −3 units/hour. Net = +1 unit/hour, so the empty tank fills in 36 hours. The drain cancels most of the inlet's work, which is why the time is so long.

[[fig:0]]

For one inlet (x hours) and one outlet (y hours):

• If x < y (inlet faster), the tank fills in xyy − x.
• If y < x (outlet faster), a full tank empties in xyx − y.

With 9 and 12: 1083 = 36 hours.

Read the starting condition carefully. Three situations come up:

• The tank is empty and the net is positive ⇒ time to fill = capacitynet.
• The tank is full (or partly full) and the net is negative ⇒ time to empty = water present|net|. If the tank is 14 full, only 14 of the capacity has to be removed.
• Many identical inlets and outlets ⇒ multiply each efficiency by the number of pipes, then take the net.

[[fig:1]]

Finding a missing pipe. If the combined time is given and one pipe is unknown, find the combined efficiency from the given time, then subtract the known pipes with their signs. What is left is the unknown pipe; if it comes out negative, that pipe is an outlet.

[[fig:2]]

A well-known question: "Normally the tank fills in T hours, but because of a leak it takes t hours more. In how much time will the leak empty the full tank?"

Without the leak, the inlets fill 1T of the tank per hour. With the leak, the tank fills in (T + t) hours, so the net filling is 1T + t per hour. The difference between the two is exactly the leak's work:

Leak = 1T − 1T + t = tT(T + t) ⇒ the leak empties the full tank in T(T + t)t

[[fig:0]]

Radhika's tank normally fills in 4 hours, but a crack makes it take 5 hours. With capacity = LCM(4, 5) = 20 units, the pipe gives +5 units/hour and the net is +4 units/hour, so the crack takes out 1 unit/hour and alone would empty the full tank in 20 hours. The formula agrees: 4 × 51 = 20 ✓.

[[fig:1]]

When two pipes are filling, first find their time together (T = xyx + y), then apply the same idea. Keep the extra time t in the same unit as T: "1 hour 20 minutes more" is 113 hours, and "45 minutes more" is 34 hour.

In these questions the story has two or more phases. In each phase a fixed set of pipes is running. The method does not change: work of phase 1 + work of phase 2 + … = capacity.

So you move forward in time:

• Find the work done in the first phase = (net efficiency of the pipes running) × (time).
• Subtract it from the capacity to get the remaining part.
• Divide the remaining part by the net efficiency of the pipes running in the next phase.

[[fig:0]]

Radhika's tank has pipe A (8 hours) and pipe B (24 hours). Both are opened, and after 3 hours A is closed. Capacity = 24 units; A = 3, B = 1. In 3 hours they fill 3 × 4 = 12 units. B alone fills the remaining 12 units in 12 hours. Total time = 3 + 12 = 15 hours.

[[fig:1]]

The same frame handles every variation:

• Pipe opened later (an outlet opened after 1 hour, a second pipe at 9 a.m.): phase 1 has fewer pipes.
• Pipe closed or leak repaired after some time: phase 2 has fewer pipes.
• Rate changes ("A's rate doubles, B's becomes half"): phase 2 uses the new efficiencies.
• Unknown pipe: write the phase equation and solve for the one unknown efficiency.
• Pairs given (A+B, B+C, C+A): first find each single pipe's units, then run the phases.

Before you mark the answer, see what the last line asks: the total time, the time after the change, or a clock time. A wrong option is often one of the other two.

Pipes opened one after another. "A is opened first, B one hour later and C one hour after that" is just a chain of phases. Radhika's tank of 30 units has A = 3, B = 2 and C = 1 units/hour. The first hour gives 3 units, the second hour (A + B) gives 5 units, and from then on all three give 6 units an hour. The remaining 30 − 8 = 22 units take 226 = 323 hours, so the tank is full 523 hours (5 h 40 min) after A was opened. When the total time is known and a pipe is missing, write each pipe's working hours instead: A works T hours, B works (T − 1) and C works (T − 2), and their work adds up to the capacity.

[[fig:2]]

This looks like Type 6, with one big difference. In Type 6 we know when the change happens (after 3 hours, at 9 a.m.), so we move forward. Here the change is fixed from the end ("A was closed 5 hours before the tank was full"), or the finishing time is fixed and we have to find when to close a pipe. So we work backward.

Start from this fact: the pipe that is never closed works for the whole time. Let the total time be T.

• A pipe closed t hours before the end works for (T − t) hours.
• Work of every pipe = its efficiency × the hours it actually ran.
• Total of all these works = capacity. Solve for T.

[[fig:0]]

Radhika's tank has pipe A (12 hours) and pipe B (4 hours). Both start together, and B is closed 2 hours before the tank is full. Capacity = 12 units; A = 1, B = 3. A runs T hours, B runs (T − 2) hours:
T + 3(T − 2) = 12 ⇒ 4T = 18 ⇒ T = 412 hours.

[[fig:1]]

The reverse question, "after how much time should B be closed so that the tank fills in exactly T hours?", is easier still. A runs for all T hours; find the part A fills, and whatever is left was filled by B. That gives B's running time directly.

An outlet closed before the end works the same way; it only carries a minus sign in the equation.

"The pipes are opened on alternate hours, starting with A." Only one pipe works at a time, so we cannot add efficiencies directly. Instead we make a cycle: A for one hour, then B for one hour. Work of one cycle (2 hours) = A + B.

The method:

• Find the work of one cycle.
• Take as many complete cycles as possible without crossing the capacity.
• Finish the remaining work with the pipe whose turn comes next: first A, and if that is not enough, then B.

Radhika's tank has pipe A (6 hours) and pipe B (9 hours), opened alternately for an hour each, starting with A. Capacity = 18 units; A = 3, B = 2; one cycle (2 hours) = 5 units. 3 cycles = 15 units in 6 hours. The remaining 3 units are exactly A's next hour. Total = 7 hours.

[[fig:0]]

Inlet and outlet alternating. This is where most students go wrong. Suppose the inlet gives +4 and the outlet −3 per hour, with capacity 12 units. One cycle gives only +1, so many write "12 cycles = 24 hours". That is wrong. The tank gets full during an inlet hour, before the outlet gets its turn. So keep one inlet turn aside first: complete cycles must bring the tank to 12 − 4 = 8 units. That takes 8 cycles = 16 hours, and the inlet's next hour fills the last 4 units. Answer = 17 hours.

[[fig:1]]

So for inlet–outlet alternation, complete cycles are needed only up to (capacity − inlet's one turn), and then the inlet finishes.

So far the tank was "1 tank" or "LCM units". Now the question brings real litres or gallons. There are only three situations.

1. Rates are given in litres. "A pours 5 litres every 2 minutes", or "A fills a 240-litre tank in 212 minutes". Convert each to litres per minute, add them, and divide the required litres by the total rate. A handy alternative: take a block of time equal to the LCM of the given minutes and count the litres in that block.

[[fig:0]]

2. One pipe is given in litres, the others in time. The standard form: "Two pipes fill the tank in 12 and 15 minutes; a waste pipe drains 10 litres per minute; with all three open, the tank fills in 20 minutes. Find the capacity." Use LCM units for everything given in time.
Say this is Radhika's tank: capacity = LCM(12, 15, 20) = 60 units. A = 5, B = 4, net with the waste pipe = 6020 = 3. So the waste pipe = 5 + 4 − 3 = 6 units/min. It drains 10 litres/min, so 6 units = 10 litres ⇒ 1 unit = 53 litres. Capacity = 60 × 53 = 100 litres.
Find the unknown pipe in units, match it with its litres, and you know the size of one unit.

[[fig:1]]

3. Changing the container size. Number of buckets × capacity of one bucket = capacity of the tank. If each bucket becomes smaller, more buckets are needed, in the inverse ratio. And when a simple fraction of the water drains out ("27 drains out"), what is left is the other part of the same amount.

[[fig:2]]

Overflow. A tank cannot hold more than its capacity. If a tap keeps running after the tank is full, the extra water spills out. So first find the empty space = capacity − water already present. Water poured beyond that space is the overflow. For example, a 300-litre tank already has 90 litres, and a tap pouring 20 litres a minute is left on for 18 minutes. It pours 360 litres, the space is only 210 litres, so 150 litres overflow. If the question asks when the overflow starts, divide the empty space by the net rate of filling.

[[fig:3]]