Mixture and Alligation A vessel has 30l of wine and 10l of water mixture . 4l of mixture is taken out from vessel and 4l of water is added.This process is repeated one more time.the ratio of quantity of wine is left and initial quantity of wine is : I have tried: Total quantity of vessel is: 40l that is: 3:1 first time :4l of mixture is taken out 3x-3/4*4 : x-1/4*4 + 4 ( this four is water is added) so ratio becomes: 3x-3:x-3 second time process is continued : (3x-3) -(3x-3)*4/4x-6 :(x-1) -(x-1)*4/4x-6 + 4 (four water is added) for second time process i have framed the equation: (3x-3) -(3x-3)*4/4x-6 + (x-1) -(x-1)*4/4x-6 + 4 = 4x - 2 x^2 is coming so please anyone guide me for what i am doing the mistake in framing the equation and solve it please

> Alternative Method -

Ratio = $\left( 1 - \frac{a}{x + y} \right)^n$

a is amount of water added.

x is quantity of wine.

y is quantity of water.

You can simply say x + y total solution.

n is number of time process repeated.

= $\left( 1 - \frac{4}{30 + 10} \right)^2$

= $\left( 1 - \frac{4}{40} \right)^2$

On solving you get 81:100