A cistern is supplied by 3 pipes, two of them conveying equal volume of water. When one of these two pipes and third pipe are simultaneously opened the cistern is filled in 12 min but when all the pipes are open cistern is filled in 7minand 30sec.How long
Certainly! Below is a step-by-step solution for the problem, along with the corresponding LaTeX code.
### Steps to the Solution
1. **Define Variables**:
Let \( a \) be the time taken by the first pipe to fill the cistern, \( a \) also be the time taken by the second pipe to fill the cistern (since they have equal volume), and let \( b \) be the time taken by the third pipe to fill the cistern.
2. **Rate of Work**:
The rate of work for each pipe is as follows:
- The rate of the first pipe = \( \frac{1}{a} \) (cistern per minute)
- The rate of the second pipe = \( \frac{1}{a} \) (cistern per minute)
- The rate of the third pipe = \( \frac{1}{b} \) (cistern per minute)
3. **Combined Rates**:
- When the first pipe and the third pipe are opened, the combined rate is
\[
\frac{1}{a} + \frac{1}{b}
\]
- They fill the cistern in 12 minutes:
\[
\frac{1}{a} + \frac{1}{b} = \frac{1}{12}
\]
- When all pipes are opened, the combined rate is
\[
\frac{1}{a} + \frac{1}{a} + \frac{1}{b} = \frac{2}{a} + \frac{1}{b}
\]
- They fill the cistern in 7 minutes and 30 seconds (which is \( \frac{15}{2} \) minutes):
\[
\frac{2}{a} + \frac{1}{b} = \frac{2}{15}
\]
4. **Setting Up the Equations**:
We now have two equations:
\[
\frac{1}{a} + \frac{1}{b} = \frac{1}{12} \tag{1}
\]
\[
\frac{2}{a} + \frac{1}{b} = \frac{2}{15} \tag{2}
\]
5. **Solve the System of Equations**:
We will solve these equations to find \( a \) and \( b \).
### LaTeX Code
Here is the LaTeX code for the entire solution including the preamble:
```latex
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\title{Cistern Problem Solution}
\author{}
\date{}
\maketitle
\section*{Problem Statement}
A cistern is supplied by 3 pipes, two of them conveying equal volume of water. When one of these two pipes and the third pipe are simultaneously opened, the cistern is filled in 12 minutes, but when all the pipes are open, the cistern is filled in 7 minutes and 30 seconds. We need to find the time taken by each pipe to fill the cistern separately.
\subsection*{Solution}
Let \( a \) be the time taken by the first pipe to fill the cistern, and since the second pipe has equal volume, it also takes \( a \) time. Let \( b \) be the time taken by the third pipe.
\subsubsection*{Flow Rates}
The rates of work for the pipes are:
\[
\text{Rate of first pipe} = \frac{1}{a}, \quad \text{Rate of second pipe} = \frac{1}{a}, \quad \text{Rate of third pipe} = \frac{1}{b}
\]
\subsubsection*{Equation for one pipe and the third}
When the first pipe and the third pipe are opened together:
\[
\frac{1}{a} + \frac{1}{b} = \frac{1}{12}
\tag{1}
\]
\subsubsection*{Equation for all pipes}
When all pipes are opened together:
\[
\frac{2}{a} + \frac{1}{b} = \frac{2}{15}
\tag{2}
\]
\subsection*{Solving the equations}
From equation (1):
\[
\frac{1}{b} = \frac{1}{12} - \frac{1}{a} \implies b = \frac{12a}{a - 12}
\]
Substituting \( \frac{1}{b} \) in equation (2):
\[
\frac{2}{a} + \left( \frac{1}{12} - \frac{1}{a} \right) = \frac{2}{15}
\]
This simplifies to:
\[
\frac{1}{a} + \frac{1}{12} = \frac{2}{15}
\]
Multiplying through by \( 60a \):
\[
60 + 5a = 8a \implies 3a = 60 \implies a = 20
\]
Finding \( b \) from \( a \):
\[
\frac{1}{b} = \frac{1}{12} - \frac{1}{20}
\]
Calculating \( \frac{5}{60} - \frac{3}{60} = \frac{2}{60} \):
\[
b = 30
\]
\subsection*{Conclusion}
Thus, the time taken by each pipe to fill the cistern is:
\begin{itemize}
\item First Pipe: 20 minutes
\item Second Pipe: 20 minutes
\item Third Pipe: 30 minutes
\end{itemize}
\end{document}
```
You can compile this code in a LaTeX editor or online LaTeX editor to see the formatted output.
Update (2024-07-27):
To solve this problem, let's denote the rates of the pipes involved:
- Let the rate of each of the two equal pipes be \( r \) (in cisterns per minute).
- Let the rate of the third pipe be \( s \) (in cisterns per minute).
From the problem, we are given two scenarios.
### Scenario 1:
When one of the equal pipes and the third pipe are opened, the cistern is filled in 12 minutes. Therefore, the combined rate of one equal pipe and the third pipe is:
\[
r + s = \frac{1}{12} \quad \text{(in cisterns per minute)}
\]
### Scenario 2:
When all three pipes are opened, the cistern is filled in 7 minutes and 30 seconds. Converting this time into minutes gives us:
\[
7 \text{ min } 30 \text{ sec } = 7.5 \text{ min}
\]
Thus, the combined rate of all three pipes is:
\[
r + r + s = 2r + s = \frac{1}{7.5} = \frac{2}{15} \quad \text{(in cisterns per minute)}
\]
### Setting up the equations
We now have the following system of equations:
1. \( r + s = \frac{1}{12} \)
2. \( 2r + s = \frac{2}{15} \)
### Solving the equations
We can solve these equations step by step.
From the first equation, we can express \( s \) in terms of \( r \):
\[
s = \frac{1}{12} - r
\]
Now, substituting this expression for \( s \) into the second equation:
\[
2r + \left(\frac{1}{12} - r\right) = \frac{2}{15}
\]
This simplifies to:
\[
2r - r + \frac{1}{12} = \frac{2}{15}
\]
\[
r + \frac{1}{12} = \frac{2}{15}
\]
To isolate \( r \), we can subtract \( \frac{1}{12} \) from both sides:
\[
r = \frac{2}{15} - \frac{1}{12}
\]
To perform this subtraction, we need a common denominator for \( 15 \) and \( 12 \), which is \( 60 \):
\[
\frac{2}{15} = \frac{8}{60}, \quad \frac{1}{12} = \frac{5}{60}
\]
Thus,
\[
r = \frac{8}{60} - \frac{5}{60} = \frac{3}{60} = \frac{1}{20}
\]
We have found \( r \). Now we can find \( s \):
\[
s = \frac{1}{12} - r = \frac{1}{12} - \frac{1}{20}
\]
Again finding a common denominator (which is \( 60 \)):
\[
\frac{1}{12} = \frac{5}{60}, \quad \frac{1}{20} = \frac{3}{60}
\]
Thus,
\[
s = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30}
\]
### Conclusion
Now we have \( r = \frac{1}{20} \) and \( s = \frac{1}{30} \).
To find the time taken by each pipe to fill the cistern:
- Time taken by each of the equal pipes (Pipe 1 and Pipe 2):
\[
\text{Time} = \frac{1}{\text{Rate}} = \frac{1}{\left(\frac{1}{20}\right)} = 20 \text{ minutes}
\]
- Time taken by the third pipe (Pipe 3):
\[
\text{Time} = \frac{1}{\left(\frac{1}{30}\right)} = 30 \text{ minutes}
\]
### LaTeX Code
Here is the LaTeX code representing the solution:
```latex
\documentclass{article}
\usepackage{amsmath}
\begin{document}
To solve this problem, let's denote the rates of the pipes involved:
\begin{itemize}
\item Let \( r \) be the rate of each of the two equal pipes (in cisterns per minute).
\item Let \( s \) be the rate of the third pipe (in cisterns per minute).
\end{itemize}
We are given two scenarios:
\textbf{Scenario 1:} One equal pipe and the third pipe opened fills the cistern in 12 minutes, so:
\[
r + s = \frac{1}{12}
\]
\textbf{Scenario 2:} All three pipes opened fill the cistern in 7 minutes and 30 seconds (\( 7.5 \) minutes), thus:
\[
2r + s = \frac{2}{15}
\]
Now we have the system of equations:
1. \( r + s = \frac{1}{12} \)
2. \( 2r + s = \frac{2}{15} \)
From the first equation, we express \( s \):
\[
s = \frac{1}{12} - r
\]
Substituting in the second equation:
\[
2r + \left(\frac{1}{12} - r\right) = \frac{2}{15}
\]
Simplifying:
\[
r + \frac{1}{12} = \frac{2}{15}
\]
Finding a common denominator (60):
\[
\frac{2}{15} = \frac{8}{60}, \quad \frac{1}{12} = \frac{5}{60}
\]
So we have:
\[
r = \frac{8}{60} - \frac{5}{60} = \frac{3}{60} = \frac{1}{20}
\]
Now we find \( s \):
\[
s = \frac{1}{12} - r = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30}
\]
Thus:
\begin{itemize}
\item Time for each of the equal pipes: \( \frac{1}{\left(\frac{1}{20}\right)} = 20 \) minutes.
\item Time for the third pipe: \( \frac{1}{\left(\frac{1}{30}\right)} = 30 \) minutes.
\end{itemize}
\end{document}
```
Copy this code into a `.tex` file to compile and display the solution using a LaTeX editor.