In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3. If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?

Quantity of alohol in the mixture = 40 x 5/8 = 25 lit

Quantity of water = 40 - 25 = 15 lit

According to question,

Required ratio =  $\frac{20 - 40 x \frac{20}{100} x \frac{5}{8}}{15 - 40 x \frac{20}{100} x \frac{3}{8} + 40 x \frac{20}{100}} = \frac{20}{15 - 3 + 8} = 1 : 1$