Question 936149
initial quantity of mixture  = 400 L
alcohol %  = 80 %
quantity of alcohol in the mixture  =  80 % of 400 L
                                  ={{{  (80/100)*400}}}
                                  ={{{80*4}}}
                              ={{{320}}} L
Let X be the quantity of water added to the mixture
        final quantity of the mixture  = 400+ x
alcohol 5 in final mixture  = 60 %
quantity of alcohol in final mixture  = 60 % of (400+x)
                                ={{{ (60/100)*(400+x)}}}
                          = {{{(3/5)*(400+x)}}}

quantity of alcohol in the initial mixture and final mixture is same
        {{{ 320  =  (3/5)*(400+x)}}}
          multiply with 5 on both sides
 {{{320 *5 = (3/5)*(400+x)*5}}}
           {{{1600 = 3*(400+x)}}}
        {{{1600 = 3*400+3*x}}}
        {{{1600 =1200+3x}}}
           move 1200 to the left
   {{{1600-1200 =3x}}}
   {{{400 =3x}}}
      divide with 3 on both sides
 {{{400/3  =3x/3}}}
      {{{400/3 = x}}}
        {{{x =133.33 L}}}
Result: Quantity of water added  = 133.33 L