Question

2 pipes running together can fill a cistern in 6 minutes if one tap takes 5 minutes more than the other to fill the cistern. Find the time in which each pipe would fill the cistern?

Answer 1

Hint: For solving these types of questions let us assume that the time in which the first pipe would fill the cistern individually is $x$ minutes. From the basic concept, we can say that the sum of the part of the cistern of each pipe individually filled in 1 minute will be equal to the part of the cistern filled by 2 pipes simultaneously in 1 minute. By performing further simplifications we will end up having the answer.

Complete step-by-step solution:
Let us assume that the time in which the first pipe would fill the cistern individually is $x$ minutes.
From the question, we have that the time in which the second pipe will fill the cistern is 5 minutes more than the one with the first pipe.
Now we can say that the time in which the second pipe will fill the cistern individually will be $x+5$ minutes.
Given in the question that 2 pipes running together can fill a cistern in 6 minutes.
So we can say that the sum of the part of the cistern of each pipe individually filled in 1 minute will be equal to the part of the cistern filled by 2 pipes simultaneously in 1 minute.
This can be mathematically expressed as
$\dfrac{1}{x}+\dfrac{1}{x+5}=\dfrac{1}{6}$
This can be further simplified as
$\dfrac{x+5+x}{x\left( x+5 \right)}=\dfrac{1}{6}$
By performing further simplifications we will have $\dfrac{2x+5}{x\left( x+5 \right)}=\dfrac{1}{6}$
This can be written as $\dfrac{2x+5}{{{x}^{2}}+5x}=\dfrac{1}{6}$
By performing L.H.S and R.H.S transformations for further simplifications, $6\left( 2x+5 \right)={{x}^{2}}+5x$
By performing further simplifications we will get
$\begin{align}
  & 12x+30={{x}^{2}}+5x \\
 & \Rightarrow {{x}^{2}}+5x-12x-30=0 \\
 & \Rightarrow {{x}^{2}}-7x-30=0 \\
 & \Rightarrow {{x}^{2}}-10x+3x-30=0 \\
 & \Rightarrow x\left( x-10 \right)+3\left( x-10 \right)=0 \\
 & \Rightarrow \left( x+3 \right)\left( x-10 \right)=0 \\
\end{align}$
From this we can conclude that $x=10\And x=-3$ are the possible values.
As here are assuming $x$ as the time in which the water from the pipe is flowing so it can’t be negative. Hence we can say that $x=10$.
Hence, we can conclude that the time in which each pipe would fill the cistern is $10\min $ and $15\min $ respectively.

Note: In work time problems like this one, the equations must be made with respect to work completed as reference. The trick in solving this question is that it requires solving expressions on a common base of LCM. And we should note that time is never negative.