# It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?

Answer

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Hint – In order to solve this problem consider the time taken from pipes of larger and smaller diameter as variables and make an equation as the given conditions. There should be minimum 2 equations, since we have to find the time of two pipes of different diameter. Doing this will give you the correct answer to this question.

Complete step-by-step answer:

Let there be two pipes A and B and the diameter of A is larger than B.

Now suppose that: Pipe A takes x hours and Pipe B takes y hours to fill the pool separately.

So we can say that:

In 1 hour pipe A can fill the pool = $\dfrac{{\text{1}}}{{\text{x}}}$

In 1 hour pipe B can fill the pool = $\dfrac{{\text{1}}}{{\text{y}}}$

It is given that if both pipes are together then they take 12 hours to fill the pool.

If the are together then the equation can be formulated as in 1 hour they fill the pool:

$\dfrac{{\text{1}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{y}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{12}}}}\,\,\,\,\,\,......(1)$

Now, in 4 hour pipe A can fill the pool is = $\dfrac{1}{{\text{x}}} \times 4 = \dfrac{4}{{\text{x}}}$.

So, in 9 hour pipe B can fill the pool is = $\frac{{{1}}}{{\text{y}}} \times 9 = \dfrac{9}{{\text{y}}}$.

Than the new equation can be formulated as they fill half of the pool the equation can be written as:

$\dfrac{4}{{\text{x}}}{\text{ + }}\dfrac{9}{{\text{y}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\,\,\,\,\,\,......(2)$

Now Let $\dfrac{{\text{1}}}{{\text{x}}}{\text{ = p}}$ and $\dfrac{{\text{1}}}{{\text{y}}}{\text{ = q}}$ ……….(3)

putting in equation (1) and (2), than the equations can be written as:

\[\begin{array}{*{20}{l}}

\begin{gathered}

{\text{p + q = }}\dfrac{{\text{1}}}{{{\text{12}}}} \\

\Rightarrow {\text{12p + 12q = 1}}\,\,\,\,\,\,......\left( 4 \right) \\

\end{gathered} \\

\begin{gathered}

\\

{\text{And}} \\

\end{gathered} \\

{{\text{4p + 9q = }}\dfrac{{\text{1}}}{{\text{2}}}} \\

{ \Rightarrow {\text{8p + 18q = 1}}\,\,\,\,\,\,.....\left( 5 \right)}

\end{array}\]

On solving equation (4) and (5) we will get the value of p and q.

On multiplying equation (4) by 2 and (5) by 3 we get the equation as:

\[ \Rightarrow 24{\text{p + 24q = 2}}\]

\[ \Rightarrow 24{\text{p + 54q = 3}}\]

On subtracting the above equations the term containing p will be cancelled out we get the value of q as:

-30q = -1

q = $\dfrac{1}{{30}}$

then putting the value of q in equation (4) we get the value of p as:

12p = 1 - $\dfrac{2}{5}$= $\dfrac{3}{5}$

So, p = $\dfrac{1}{{20}}$

From (3) we can say that x=20 and y=30.

Hence the A pipe would take 20 hours and the pipe B would take 30 hours to fill the pool separately.

Note – Whenever you face such types of problems then assume the variables wherever needed according to the conditions provided. Then solve the equation to get the variables. Remember there must be the number of equations equal to the number of variables to get the answer. Here we have assumed $\dfrac{{\text{1}}}{{\text{x}}}{\text{ = p}}$ and $\dfrac{{\text{1}}}{{\text{y}}}{\text{ = q}}$ to make our equations easy to solve. Proceeding like this will make your problem solved.

Complete step-by-step answer:

Let there be two pipes A and B and the diameter of A is larger than B.

Now suppose that: Pipe A takes x hours and Pipe B takes y hours to fill the pool separately.

So we can say that:

In 1 hour pipe A can fill the pool = $\dfrac{{\text{1}}}{{\text{x}}}$

In 1 hour pipe B can fill the pool = $\dfrac{{\text{1}}}{{\text{y}}}$

It is given that if both pipes are together then they take 12 hours to fill the pool.

If the are together then the equation can be formulated as in 1 hour they fill the pool:

$\dfrac{{\text{1}}}{{\text{x}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{y}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{12}}}}\,\,\,\,\,\,......(1)$

Now, in 4 hour pipe A can fill the pool is = $\dfrac{1}{{\text{x}}} \times 4 = \dfrac{4}{{\text{x}}}$.

So, in 9 hour pipe B can fill the pool is = $\frac{{{1}}}{{\text{y}}} \times 9 = \dfrac{9}{{\text{y}}}$.

Than the new equation can be formulated as they fill half of the pool the equation can be written as:

$\dfrac{4}{{\text{x}}}{\text{ + }}\dfrac{9}{{\text{y}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\,\,\,\,\,\,......(2)$

Now Let $\dfrac{{\text{1}}}{{\text{x}}}{\text{ = p}}$ and $\dfrac{{\text{1}}}{{\text{y}}}{\text{ = q}}$ ……….(3)

putting in equation (1) and (2), than the equations can be written as:

\[\begin{array}{*{20}{l}}

\begin{gathered}

{\text{p + q = }}\dfrac{{\text{1}}}{{{\text{12}}}} \\

\Rightarrow {\text{12p + 12q = 1}}\,\,\,\,\,\,......\left( 4 \right) \\

\end{gathered} \\

\begin{gathered}

\\

{\text{And}} \\

\end{gathered} \\

{{\text{4p + 9q = }}\dfrac{{\text{1}}}{{\text{2}}}} \\

{ \Rightarrow {\text{8p + 18q = 1}}\,\,\,\,\,\,.....\left( 5 \right)}

\end{array}\]

On solving equation (4) and (5) we will get the value of p and q.

On multiplying equation (4) by 2 and (5) by 3 we get the equation as:

\[ \Rightarrow 24{\text{p + 24q = 2}}\]

\[ \Rightarrow 24{\text{p + 54q = 3}}\]

On subtracting the above equations the term containing p will be cancelled out we get the value of q as:

-30q = -1

q = $\dfrac{1}{{30}}$

then putting the value of q in equation (4) we get the value of p as:

12p = 1 - $\dfrac{2}{5}$= $\dfrac{3}{5}$

So, p = $\dfrac{1}{{20}}$

From (3) we can say that x=20 and y=30.

Hence the A pipe would take 20 hours and the pipe B would take 30 hours to fill the pool separately.

Note – Whenever you face such types of problems then assume the variables wherever needed according to the conditions provided. Then solve the equation to get the variables. Remember there must be the number of equations equal to the number of variables to get the answer. Here we have assumed $\dfrac{{\text{1}}}{{\text{x}}}{\text{ = p}}$ and $\dfrac{{\text{1}}}{{\text{y}}}{\text{ = q}}$ to make our equations easy to solve. Proceeding like this will make your problem solved.

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