Running together can fill time 11 …

By the question,two pipes running together can fill a tank in {tex}11 \frac { 1 } { 9 }{/tex} minutes. If one pipe takes 5 minutes more  than the other to fill the tank,we have to find the time in which each pipe would fill the tank separately.

Let time taken by pipe A be x minutes, and time taken by pipe B be x + 5 minutes.

In one minute pipe A will fill {tex}\frac { 1 } { x }{/tex} tank

In one minute pipe B will fill {tex}\frac { 1 } { x + 5 }{/tex} tank

pipes A + B will fill in one minute = {tex}\frac { 1 } { x } + \frac { 1 } { x + 5 }{/tex} tank

Now according to the question.

{tex}\frac { 1 } { x } + \frac { 1 } { x + 5 } = \frac { 9 } { 100 }{/tex}

or, {tex}\frac { x + 5 + x } { x ( x + 5 ) } = \frac { 9 } { 100 }{/tex}

or, 100(2x + 5) = 9x(x + 5)

or, {tex}200 x + 500 = 9 x ^ { 2 } + 45 x{/tex}

or, {tex}9 x ^ { 2 } - 155 x - 500 = 0{/tex}

or, {tex}9 x ^ { 2 } - 180 x + 25 x - 500 = 0{/tex}

or,9x(x - 20) + 25(x - 20) = 0

or, (x-20)(9x + 25) = 0

or, {tex}x = 20 , \frac { - 25 } { 9 }{/tex}

rejecting negative value, x = 20 minutes

and  x + 5 = 25 minutes 

Hence pipe A will fill the tank in 20 minutes and pipe B will fill it in 25 minutes.