The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture?

           Wine                         Water

            8L                              32L

            1                :               4

           20 %                           80% (original ratio)

           30 %                           70% (required ratio)

In  ths case, the percentage of water being reduced when the mixture is being replaced with wine.

so the ratio of left quantity to the initial quantity is 7:8

Therefore ,  $\frac{7}{8}=1-\frac{K}{40}$

=> K = 5 Lit