Pipes and Cisterns

Pipes and Cisterns is a topic that often appears in quantitative aptitude sections of competitive exams. It's a direct application of the concepts of work and time. The core idea is to understand how fast a pipe can fill or empty a tank and how these rates combine when multiple pipes are involved.

Understanding the Basics

In work and time problems, we usually talk about people doing a job. Here, we're talking about pipes filling or emptying a tank. A tank can be thought of as the "job" to be done.

• Rate of a pipe: This is the amount of tank filled or emptied by a pipe in one unit of time (usually one hour or one minute).
• Positive rate: If a pipe fills the tank, its rate is considered positive.
• Negative rate: If a pipe empties the tank, its rate is considered negative.

Calculating Efficiency

The efficiency of a pipe is its rate of work. If a pipe can fill a tank of capacity 'C' in 'T' hours, its rate of filling is C/T. For simplicity, we often consider the capacity of the tank to be 1 unit.

If a pipe can fill a tank in x hours, its rate of filling is 1/x of the tank per hour.

If a pipe can empty a tank in y hours, its rate of emptying is 1/y of the tank per hour.

Combined Work of Pipes

When multiple pipes are working together, their rates are added or subtracted depending on whether they are filling or emptying.

Let Pipe A fill the tank in x hours and Pipe B fill the tank in y hours. If both pipes are opened together, the time taken to fill the tank is:

Rate of A = 1/x tank/hour
Rate of B = 1/y tank/hour
Combined Rate = (1/x + 1/y) tank/hour
Time taken together = 1 / (1/x + 1/y) hours = xy / (x + y) hours

If Pipe A fills the tank in x hours and Pipe B empties it in y hours.

Rate of A (filling) = 1/x tank/hour
Rate of B (emptying) = -1/y tank/hour
Combined Rate = (1/x - 1/y) tank/hour
Time taken to fill = 1 / (1/x - 1/y) hours = xy / (y - x) hours (This is valid only if y > x, meaning A fills faster than B empties).

Example Calculation

Pipe A can fill a tank in 10 hours. Pipe B can fill the same tank in 15 hours. How long will it take for both pipes to fill the tank if opened together?

Rate of Pipe A = 1/10 tank/hour
Rate of Pipe B = 1/15 tank/hour
Combined Rate = 1/10 + 1/15
To add these fractions, find a common denominator, which is 30.
Combined Rate = (3/30) + (2/30) = 5/30 = 1/6 tank/hour
Time taken together = 1 / (1/6) = 6 hours.

Example with an Outlet Pipe

Pipe A can fill a tank in 8 hours. Pipe B can empty the same tank in 12 hours. If both pipes are opened simultaneously, how long will it take to fill the tank?

Rate of Pipe A (filling) = 1/8 tank/hour
Rate of Pipe B (emptying) = -1/12 tank/hour
Combined Rate = 1/8 - 1/12
Common denominator is 24.
Combined Rate = (3/24) - (2/24) = 1/24 tank/hour
Time taken to fill = 1 / (1/24) = 24 hours.

Dealing with Complex Scenarios

Sometimes, pipes are opened at different times, or one pipe is closed after some time. In such cases, calculate the work done by each pipe during its operational period and then find the remaining work.

Scenario: Pipe A fills a tank in 20 hours. Pipe B empties it in 30 hours. Pipe A is opened for the first 10 hours, then Pipe B is also opened. How long will it take to fill the tank from the moment Pipe B is opened?

Rate of A = 1/20 tank/hour
Rate of B = -1/30 tank/hour

Work done by A in the first 10 hours = Rate of A * Time = (1/20) * 10 = 10/20 = 1/2 tank.

Remaining work to be filled = 1 - 1/2 = 1/2 tank.

Now, both A and B are open.
Combined Rate = Rate of A + Rate of B = (1/20) + (-1/30) = 1/20 - 1/30
Common denominator is 60.
Combined Rate = (3/60) - (2/60) = 1/60 tank/hour.

Time taken to fill the remaining 1/2 tank = Remaining Work / Combined Rate
Time = (1/2) / (1/60) = (1/2) * 60 = 30 hours.

So, it will take 30 hours from the moment Pipe B is opened to fill the tank.

Time Speed and Distance (TSD)

Time Speed and Distance is a fundamental topic in quantitative aptitude. It deals with the relationship between three quantities: Time, Speed, and Distance. The basic formula connects these three, and problems often involve variations of this formula, relative speeds, or average speeds.

The Fundamental Formula

The core relationship is:

Distance = Speed × Time

From this, we can derive:

• Speed = Distance / Time
• Time = Distance / Speed

Units of Measurement

It is crucial to maintain consistent units.

• If distance is in kilometers (km) and time is in hours (hr), speed will be in kilometers per hour (km/hr).
• If distance is in meters (m) and time is in seconds (sec), speed will be in meters per second (m/s).

Unit Conversions

Common conversions needed:

• km/hr to m/s: To convert km/hr to m/s, multiply by 5/18.
  1 km/hr = 1000 meters / 3600 seconds = 10/36 m/s = 5/18 m/s.
• m/s to km/hr: To convert m/s to km/hr, multiply by 18/5.
  1 m/s = (1/1000) km / (1/3600) hr = 3600/1000 km/hr = 36/10 km/hr = 18/5 km/hr.

Average Speed

Average speed is not simply the average of the speeds. It is defined as the total distance traveled divided by the total time taken.

Average Speed = Total Distance / Total Time

Case 1: Equal Distances
If an object travels a distance 'd' at speed 's1' and then travels the same distance 'd' at speed 's2', the average speed is:
Time taken for first part = d/s1
Time taken for second part = d/s2
Total Distance = d + d = 2d
Total Time = d/s1 + d/s2 = d(1/s1 + 1/s2) = d((s1+s2)/s1*s2)
Average Speed = 2d / [d((s1+s2)/s1*s2)] = 2 / [(s1+s2)/s1*s2] = 2 * s1 * s2 / (s1 + s2) (Harmonic Mean)

Case 2: Equal Times
If an object travels for time 't' at speed 's1' and then travels for the same time 't' at speed 's2', the average speed is:
Distance covered in first part = s1 * t
Distance covered in second part = s2 * t
Total Distance = s1*t + s2*t = t(s1 + s2)
Total Time = t + t = 2t
Average Speed = t(s1 + s2) / 2t = (s1 + s2) / 2 (Arithmetic Mean)

Example of Average Speed

A car travels from City A to City B at a speed of 40 km/hr and returns from City B to City A at a speed of 60 km/hr. What is the average speed for the entire journey?

Here, the distance is the same for both legs of the journey. So we use the formula for equal distances:
Average Speed = 2 * s1 * s2 / (s1 + s2)
Average Speed = 2 * 40 * 60 / (40 + 60)
Average Speed = 2 * 40 * 60 / 100
Average Speed = 4800 / 100 = 48 km/hr.

Note that the average speed (48 km/hr) is less than the simple average of 40 and 60 (which is 50 km/hr). This is because the car spends more time traveling at the slower speed.

Relative Speed

Relative speed is the speed of one object with respect to another. This concept is crucial when dealing with problems involving two objects moving simultaneously.

Case 1: Objects moving in the same direction
If two objects A and B are moving in the same direction with speeds s1 and s2 respectively (s1 > s2), the relative speed of A with respect to B is (s1 - s2). This is the speed at which the distance between them changes.

Case 2: Objects moving in opposite directions
If two objects A and B are moving towards each other or away from each other (in opposite directions) with speeds s1 and s2, their relative speed is (s1 + s2). This is the speed at which the distance between them decreases (if moving towards) or increases (if moving away).

Example of Relative Speed

Two trains, A and B, start at the same time from points P and Q towards each other. Train A travels at 50 km/hr and Train B travels at 70 km/hr. The distance between P and Q is 600 km. In how much time will they meet?

Since they are moving towards each other, their relative speed is the sum of their speeds.
Relative Speed = Speed of A + Speed of B = 50 km/hr + 70 km/hr = 120 km/hr.
The total distance to be covered by their combined movement is 600 km.
Time to meet = Total Distance / Relative Speed
Time = 600 km / 120 km/hr = 5 hours.

Boats and Streams

Boats and Streams problems are a specific application of Time Speed and Distance concepts. They involve a boat moving in a river, where the river's current affects the boat's speed.

Key Concepts

There are two main speeds to consider:

• Speed of the boat in still water: This is the speed the boat would have if there were no current (let's denote this as 'b').
• Speed of the stream (current): This is the speed of the river's flow (let's denote this as 's').

Speed Upstream and Downstream

When a boat moves in a river, its speed relative to the ground changes based on the direction of the river flow.

• Downstream: When the boat moves in the same direction as the current (i.e., with the flow).
  Speed Downstream (S_d) = Speed of Boat + Speed of Stream = b + s
• Upstream: When the boat moves in the opposite direction to the current (i.e., against the flow).
  Speed Upstream (S_u) = Speed of Boat - Speed of Stream = b - s

It's important to note that for upstream travel, the speed of the boat in still water (b) must be greater than the speed of the stream (s), otherwise, the boat cannot move against the current.

Formulas Derived from TSD

Using the basic TSD formula (Distance = Speed × Time), we can derive formulas for boats and streams:

• Distance Downstream = S_d × Time Downstream = (b + s) × T_d
• Distance Upstream = S_u × Time Upstream = (b - s) × T_u

If the distance traveled downstream is the same as the distance traveled upstream, then:
(b + s) × T_d = (b - s) × T_u

Finding Boat Speed and Stream Speed

If you know the downstream speed (S_d) and upstream speed (S_u), you can find the speed of the boat and the stream:

We have:
b + s = S_d (Equation 1)
b - s = S_u (Equation 2)

Adding Equation 1 and Equation 2:
(b + s) + (b - s) = S_d + S_u
2b = S_d + S_u
b = (S_d + S_u) / 2

Subtracting Equation 2 from Equation 1:
(b + s) - (b - s) = S_d - S_u
2s = S_d - S_u
s = (S_d - S_u) / 2

Example Calculation

A boat travels downstream at 15 km/hr and upstream at 5 km/hr. Find the speed of the boat in still water and the speed of the stream.

Speed Downstream (S_d) = 15 km/hr
Speed Upstream (S_u) = 5 km/hr

Speed of boat (b) = (S_d + S_u) / 2 = (15 + 5) / 2 = 20 / 2 = 10 km/hr.
Speed of stream (s) = (S_d - S_u) / 2 = (15 - 5) / 2 = 10 / 2 = 5 km/hr.

Another Example

A man can row a boat at a speed of 10 km/hr in still water. If the river is flowing at 4 km/hr, how much time will he take to travel 42 km downstream and then return to the starting point?

Speed of boat in still water (b) = 10 km/hr
Speed of stream (s) = 4 km/hr

Speed Downstream (S_d) = b + s = 10 + 4 = 14 km/hr.
Speed Upstream (S_u) = b - s = 10 - 4 = 6 km/hr.

Distance = 42 km (for both downstream and upstream journeys).

Time taken to travel downstream = Distance / S_d = 42 km / 14 km/hr = 3 hours.
Time taken to travel upstream = Distance / S_u = 42 km / 6 km/hr = 7 hours.

Total time taken for the round trip = Time Downstream + Time Upstream = 3 hours + 7 hours = 10 hours.