Question

If two pipes function simultaneously, the reservoir will be filled in $12$ hours. Through one pipe the reservoir is filled up $10$ hours after than through the other. How many hours does it take the second pipe to fill the reservoir?

Answer 1

Hint: From the question, we are going to find how much time taken by the second pipe to fill the reservoir alone. By the given, we know that the two pipes function simultaneously to fill the reservoir in $12$ hours.
Pipes and cistern is another format of time and work based problems. The problems based on pipes and cistern is a common topic from which questions are asked in the quantitative aptitude sections for various competitive exams.

Formula used: If a pipe can fill a tank in ${\text{x}}$ hours, then a part of the tank is filled in $1$hour$ = \dfrac{1}{{\text{x}}}$.

Complete step-by-step answer:
Let us consider the first pipe as a faster pipe be ${\text{A}}$.
Let us consider the second pipe as a slower pipe be ${\text{B}}$.
By the given, the two pipes ${\text{A}}$ and ${\text{B}}$ are flowing simultaneously then the reservoir will be filled in$12$ hours.
Let the faster pipe ${\text{A}}$ fill the reservoir alone in ${\text{x}}$ hour.
Let the faster pipe ${\text{A}}$ fill the reservoir in $10$ hours faster than through the slower pipe ${\text{B}}$.
Thus, the slower pipe ${\text{B}}$ fills the reservoir in \[\left( {{\text{x}} + 10} \right)\] hours.
In $1$ hour, the faster pipe ${\text{A}}$ will fill the portion of the reservoir in $\dfrac{1}{{\text{x}}}$ hour.
In $12$ hours, the faster pipe ${\text{A}}$ will fill the portion of the reservoir in $12 \times \left( {\dfrac{1}{{\text{x}}}} \right) = \dfrac{{12}}{{\text{x}}}$ hours.
In $1$ hour, the slower pipe ${\text{B}}$ will fill the portion of the reservoir in $\dfrac{1}{{\left( {{\text{x}} + 10} \right)}}$ hour.
In $12$ hours, the slower pipe ${\text{B}}$ will fill the portion of the reservoir in $12 \times \left( {\dfrac{1}{{\left( {{\text{x}} + 10} \right)}}} \right) = \dfrac{{12}}{{\left( {{\text{x}} + 10} \right)}}$ hours.
In $12$ hours, the faster pipe ${\text{A}}$ and the slower pipe ${\text{B}}$ will fill the whole portion of the reservoir equal to $1$ .
Now, express the above term in mathematical form as $\dfrac{{12}}{{\text{x}}} + \dfrac{{12}}{{{\text{x}} + 10}} = 1$ .
Now, we have to find the ${\text{x}}$ value from the above mathematical expression by using Least Common Multiple and factorization process then after some addition and subtraction, we get the ${\text{x}}$ value.
$\dfrac{{12}}{{\text{x}}} + \dfrac{{12}}{{{\text{x}} + 10}} = 1$
While taking $12$ as a common term in the numerator of left hand side (LHS). We get,
  $ \Rightarrow 12\left( {\dfrac{1}{{\text{x}}} + \dfrac{1}{{{\text{x}} + 10}}} \right) = 1$
Now, cross multiply $12$ into the right hand side (RHS).
$ \Rightarrow \dfrac{1}{{\text{x}}} + \dfrac{1}{{{\text{x}} + 10}} = \dfrac{1}{{12}}$
Now, using the Least Common Multiple (LCM) method. The LCM of ${\text{x}}$and ${\text{x}} + 10$ is ${\text{x}}\left( {{\text{x}} + 10} \right)$. We get,
$ \Rightarrow \dfrac{{{\text{x}} + 10}}{{{\text{x}}\left( {{\text{x}} + 10} \right)}} + \dfrac{{\text{x}}}{{\left( {{\text{x}} + 10} \right)\left( {\text{x}} \right)}} = \dfrac{1}{{12}}$
$ \Rightarrow \dfrac{{\left( {{\text{x}} + 10} \right) + {\text{x}}}}{{{\text{x}}\left( {{\text{x}} + 10} \right)}} = \dfrac{1}{{12}}$
$ \Rightarrow \dfrac{{2{\text{x}} + 10}}{{{\text{x}}\left( {{\text{x}} + 10} \right)}} = \dfrac{1}{{12}}$
Now, cross multiply the above term. We get,
$ \Rightarrow 12\left( {2{\text{x}} + 10} \right) = 1\left( {{\text{x}}\left( {{\text{x}} + 10} \right)} \right)$
$ \Rightarrow 24{\text{x}} + 120 = {\text{x}}\left( {{\text{x}} + 10} \right)$
$ \Rightarrow 24{\text{x}} + 120 = {{\text{x}}^2} + 10{\text{x}}$
Now, we have to arrange all terms on one side and to get a quadratic equation. We get,
$ \Rightarrow {{\text{x}}^2} + 10{\text{x}} - 24{\text{x}} - 120 = 0$
$ \Rightarrow {{\text{x}}^2} - 14{\text{x}} - 120 = 0$
By splitting the ${\text{x}}$ co-ordinates (middle term) of the quadratic equation. We get,
$ \Rightarrow {{\text{x}}^2} + 6{\text{x}} - 20{\text{x}} + 120 = 0$
Now, we have to make a common term in the quadratic equation. We get,
$ \Rightarrow {\text{x}}\left( {{\text{x}} + 6} \right) - 20\left( {{\text{x}} + 6} \right) = 0$
$ \Rightarrow \left( {{\text{x}} + 6} \right)\left( {{\text{x}} - 20} \right) = 0$
Now, separate the above term equals to zero. We get,
$ \Rightarrow \left( {{\text{x}} + 6} \right) = 0$ or $\left( {{\text{x}} - 20} \right) = 0$
$ \Rightarrow {\text{x}} = - 6$ or ${\text{x}} = 20$
Here we get two values for ${\text{x}}$. Since, the time cannot be negative, so ${\text{x}} = - 6$ is not accepted.
Therefore, ${\text{x}} = 20$ be the solution.

Thus, the first pipe ${\text{A}}$ will take $\left( {{\text{x}} = } \right)20$ hours to fill the reservoir alone and the second pipe ${\text{B}}$ will take $\left( {x + 10 \Rightarrow 20 + 10 = } \right)30$ hours to fill the reservoir alone.

Note: The students may solve the above given problem easily only when mathematical operations will be correct. So, students must concentrate in the above mathematical operations and in the solving of quadratic equations. Particularly in factoring the quadratic equation. Students mostly make mistakes on those calculations, they definitely have to focus on those calculations. Linear equations in one variable can be used when we have one unknown quantity. Linear equations in two variables can be used when we have two unknown quantities.