To fill a swimming pool 2 …

Let the time taken by the pipe of larger diameter = x hours

The time taken by the pipe of smaller diameter = x + 10 hours

In 1 hour pipe of larger diameter fills  {tex}1\over x{/tex}part of the pool , So in 4 hour the pipe of larger diameter fills = {tex}4\over x{/tex}

In 1 hour pipe of smaller diameter fills {tex}1\over x + 10{/tex} part of the pool , So in 9 hour the pipe of larger diameter fills ={tex} 9\over x + 10{/tex}
 

ATQ

{tex}{4\over x} + {9\over x + 10} = {1\over 2}{/tex}

=> {tex}{4x+40+9x\over x^2+10x}= {1\over 2}{/tex}

=> {tex}x^2+10x = 26x+80{/tex}

=> {tex}x^2-16x-80=0{/tex}

=> {tex}x^2-20x+4x-80=0{/tex}

=> x(x-20)+4(x-20)= 0

=> (x-20)(x+4)=0

x  = - 4  and 20, 

But we assume x as hour , So we neglect negative term and get

x  =  20

The time taken by the pipe of larger diameter = 20 hours
And
The time taken by the pipe of smaller diameter = 20 + 10 =  30 hours