Question

Two pipes running together can fill a cistern in 2 8/11
minutes. If one pipe takes 1 minute more than the other to fill the cistern, find the time in which each pipe would fill the cistern alone

Answer 1

So here's the solution. I hope it may help you.

Answer 2

Answer:

Pipe 1 fill the cistern alone in 5 minutes and Pipe 2 fill the cistern alone in 6 minutes

Step-by-step explanation:

Let Pipe 1 fill the cistern alone in x minutes

We are given that one pipe takes 1 minute more than the other to fill the cistern

SO, Pipe 2 fill the cistern in x+1 minutes

Pipe 1's 1 minute work = $\frac{1}{x}$

Pipe 2's 1 minute work = $\frac{1}{x+1}$

They work together in 1 minute =  $\frac{1}{x}+\frac{1}{x+1}$

We are also given that Two pipes running together can fill a cistern in 2 8/11 i.e. 30/11

So, their together 1 minute work = $\frac{11}{30}$

So,   $\frac{1}{x}+\frac{1}{x+1}=\frac{11}{30}$

$\frac{2x+1}{x^2+x}=\frac{11}{30}$

$x=\frac{-6}{11},5$

S, since minutes cannot be negative

So, x = 5

So, Pipe 1 fill the cistern alone in 5 minutes and Pipe 2 fill the cistern alone in x+1 = 5+1 = 6 minutes

Hence  Pipe 1 fill the cistern alone in 5 minutes and Pipe 2 fill the cistern alone in 6 minutes