A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by the same amount of water to make the ratio of acid and water 4 : 3 is

let the quantity of mixture be 70 lt

therefore quantity of acid = $$ 70 \times \frac{80}{100} = 56 $$

after replacement the quantity of acid = $$ \frac{70 \times 4}{7} = 40 lt $$

therefore (56 - 40 = 16 lt) of acid is removed from the original mixture

let x lt of the mixture is removed and replaced by water

$$ x \times \frac{80}{100} = 16 $$

solving x = 20 lt

required part of mixture = $$ \frac{20}{70} = \frac{2}{7} $$