Question

A cistern can be filled by three pipes with uniform flow. The first two pipes operating simultaneously, fill the cistern at the same time during which the cistern is filled by the third pipe alone. The second pipe fills the cistern $5$ hours faster than the first pipe and $4$ hours slower than the third pipe. Find the time required by each pipe to fill the cistern individually.

Answer 1

Step 1: Determining the quadratic equation in $x$

Let the time taken by the first pipe to fill the cistern individually be $x$ hours.

It is given that the second pipe fills the cistern $5$ hours faster. That means it takes $(x-5)$ hours to fill the cistern individually.

And, the third pipe fills the cistern $4$ hours faster than the second pipe. That means that the third pipe takes $(x-9)$ hours to fill the cistern individually.

$$\begin{gathered} \therefore \frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9} \\ \Rightarrow \frac{x-5+x}{x(x-5)}=\frac{1}{x-9} \\ \Rightarrow (x-9)(2x-5)=x(x-5) \\ \Rightarrow x(2x-5)-9(2x-5)=x^2-5x \\ {\Rightarrow 2x}^2-5x-18x+45=x^2-5x \\ \Rightarrow x^2-18x+45=0 \end{gathered}$$

Step 2: Finding the time required by each pipe to fill the cistern individually

By factorization method

$$\begin{gathered} x^2-18x+45=0 \\ \Rightarrow x^2-15x-3x+45=0 \\ \Rightarrow x(x-15)-3(x-15)=0 \\ \Rightarrow (x-15)(x-3)=0 \\ \therefore x=3,15 \end{gathered}$$

Since the second pipe fills the cistern 5 hours faster than the first pipe.

$\mathit{\therefore }\mathit{}x\mathit{=}\mathit{15}$

For the second pipe, $x-5=15-5=10$ hours.

For the third pipe, $x-9=15-9=6$ hours.

Therefore, the first, the second and the third pipes take $15$ hours, $10$ hours and $6$ hours respectively, to fill the cistern individually.