Pipes and Cisterns

Filling and draining problems

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Pipes and Cisterns problems deal with the filling and emptying of tanks or cisterns using pipes. These problems are analogous to Time and Work problems, where inlet pipes fill the tank (positive work) and outlet pipes empty it (negative work). Understanding rates of work and combined operations is essential for solving these efficiently.

Key Formulas

If a pipe fills a tank in ’x’ hours, its filling rate = 1/x per hour\text{If a pipe fills a tank in 'x' hours, its filling rate = 1/x per hour}
If a pipe empties a tank in ’y’ hours, its emptying rate = 1/y per hour\text{If a pipe empties a tank in 'y' hours, its emptying rate = 1/y per hour}
Net rate when both work together = (1/x - 1/y) per hour (when y > x, tank fills; when x > y, tank empties)\text{Net rate when both work together = (1/x - 1/y) per hour (when y > x, tank fills; when x > y, tank empties)}
Time taken to fill/empty = 1 / Net Rate\text{Time taken to fill/empty = 1 / Net Rate}
Part of tank filled in ’t’ hours = t x Filling Rate\text{Part of tank filled in 't' hours = t x Filling Rate}
Part of tank emptied in ’t’ hours = t x Emptying Rate\text{Part of tank emptied in 't' hours = t x Emptying Rate}
If pipe A is open for ’a’ hours and pipe B for ’b’ hours: Part filled = a/x +/- b/y\text{If pipe A is open for 'a' hours and pipe B for 'b' hours: Part filled = a/x +/- b/y}
For multiple pipes: Combined rate = 1/x +/- 1/x +/- 1/x +/- ...\text{For multiple pipes: Combined rate = 1/x +/- 1/x +/- 1/x +/- ...}
Tank capacity = LCM of individual times (useful for whole number calculations)\text{Tank capacity = LCM of individual times (useful for whole number calculations)}

Key Concepts

Inlet and Outlet Pipes

An inlet pipe fills the tank and contributes positive work. An outlet pipe (or leak) empties the tank and contributes negative work. When calculating net work, inlet pipes are added and outlet pipes are subtracted. The sign convention is crucial - always verify whether the final answer represents filling or emptying.

Work Rate Concept

The fundamental principle is that work done = rate x time. For pipes, we consider the tank as 1 unit of work. A pipe's rate is the reciprocal of the time it takes to complete the work alone. For example, a pipe that fills in 5 hours works at 1/5 of the tank per hour. This rate remains constant unless stated otherwise.

Combined Operations

When multiple pipes operate simultaneously, their rates combine algebraically. If pipes A, B fill and C empties: Net Rate = 1/A + 1/B - 1/C. Positive result means filling; negative means emptying. The time to fill/empty = 1 / |Net Rate|. Always check the sign to determine the nature of work.

Pipes Operating in Sequence

Sometimes pipes operate one after another or for different durations. Calculate work done by each pipe separately and sum them. For example, if pipe A works for 2 hours and then pipe B for 3 hours: Work done = 2 x (1/A) + 3 x (1/B). The remaining work is 1 minus this sum.

Tank Capacity Method (LCM Approach)

To avoid fractions, assume tank capacity = LCM of all given times. Then calculate each pipe's hourly capacity as (LCM/time). This converts fractional rates to whole numbers, making calculations easier. For example, if pipes take 4, 6, 8 hours: Capacity = 24 units. Rates become 6, 4, 3 units/hour respectively.

Partial Operations and Leaks

When a pipe operates only partially or there's a leak, break the problem into phases. Calculate work done in each phase separately. For leak problems, treat the leak as an outlet pipe that may operate throughout or only when the tank is being filled. The key is identifying when each component is active.

Solved Examples

Problem 1:

Pipe A can fill a tank in 6 hours and Pipe B can fill the same tank in 8 hours. If both pipes are opened together, how long will it take to fill the tank?

Solution:

Step 1: Identify individual rates
Pipe A's rate = 1/6 per hour
Pipe B's rate = 1/8 per hour

Step 2: Calculate combined rate
Combined rate = 1/6 + 1/8
= (4 + 3)/24 = 7/24 per hour

Step 3: Calculate time to fill
Time = 1 / (7/24) = 24/7 hours

Step 4: Convert to mixed number
24/7 = 3 3/7 hours = 3 hours and (3/7 x 60) minutes
= 3 hours and 180/7 minutes
~= 3 hours and 25.7 minutes

Answer: 24/7 hours or approximately 3 hours 26 minutes

Problem 2:

A pipe can fill a tank in 12 hours. Due to a leak at the bottom, it takes 15 hours to fill the tank. How long will the leak take to empty the full tank?

Solution:

Step 1: Identify rates
Filling pipe rate = 1/12 per hour
Let leak empty rate = 1/x per hour

Step 2: Set up equation for combined operation
Net rate = 1/12 - 1/x

Step 3: Use given combined time
Time with leak = 15 hours
Net rate = 1/15

Step 4: Solve for leak rate
1/12 - 1/x = 1/15
1/x = 1/12 - 1/15
1/x = (5 - 4)/60 = 1/60

Step 5: Find leak time
x = 60 hours

Answer: The leak will empty the full tank in 60 hours

Problem 3:

Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both pipes are opened together but pipe B is turned off after 10 minutes, how long will it take to fill the tank completely?

Solution:

Step 1: Convert to consistent units (minutes)
Pipe A rate = 1/20 per minute
Pipe B rate = 1/30 per minute

Step 2: Calculate combined rate for first 10 minutes
Combined rate = 1/20 + 1/30 = (3 + 2)/60 = 5/60 = 1/12 per minute

Step 3: Work done in first 10 minutes
Work = 10 x (1/12) = 10/12 = 5/6 of tank

Step 4: Remaining work
Remaining = 1 - 5/6 = 1/6 of tank

Step 5: Time for A to complete remaining work
Time = (1/6) / (1/20) = 20/6 = 10/3 minutes = 3 1/3 minutes

Step 6: Total time
Total = 10 + 10/3 = 40/3 = 13 1/3 minutes

Answer: 13 1/3 minutes or 13 minutes 20 seconds

Problem 4:

Three pipes A, B, and C are attached to a tank. A and B can fill it in 4 and 6 hours respectively, while C can empty it in 12 hours. If all three pipes are opened together when the tank is empty, how long will it take to fill the tank?

Solution:

Step 1: Identify rates
Pipe A (filling) = 1/4 per hour
Pipe B (filling) = 1/6 per hour
Pipe C (emptying) = 1/12 per hour

Step 2: Calculate net rate
Net rate = 1/4 + 1/6 - 1/12

Step 3: Find common denominator (12)
= 3/12 + 2/12 - 1/12
= (3 + 2 - 1)/12 = 4/12 = 1/3 per hour

Step 4: Calculate time to fill
Time = 1 / (1/3) = 3 hours

Answer: The tank will be filled in 3 hours

Tips & Tricks

  • Always identify inlet (filling) and outlet (emptying) pipes clearly and assign appropriate signs (+ for filling, - for emptying)
  • Use the LCM method when dealing with multiple pipes - it converts fractions to whole numbers and reduces calculation errors
  • When a leak is mentioned, treat it as an outlet pipe that works continuously while filling is happening
  • For pipes operating in sequence, calculate work done by each phase separately and track the remaining work
  • Check whether the answer should be in hours, minutes, or both - convert carefully using 1 hour = 60 minutes
  • If pipes are turned on/off at different times, break the problem into time intervals and sum the work done in each
  • For 'full tank' scenarios, remember that work completed = 1 (full tank). Partial fillings are represented as fractions
  • When comparing pipe efficiencies, the pipe with smaller time value is more efficient (works faster)
  • Verify your answer by working backwards - check if the calculated time with given rates actually fills/empties the tank

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