A $12$ -liter solution of acid and water contains $30\%$ acid. How much water (in liters) must be added to get a solution containing $20\%$ acid?

Given:
* Initial volume of the solution $=12\text{liters}$
* Initial concentration of acid $=30\%$

Let's find the volume of pure acid in the initial solution:

$$\text{Volume of Acid}=12\times \frac{30}{100}=3.6\text{liters}$$

Since only water is added to the mixture, the total volume of pure acid remains unchanged at $3.6\text{liters}$.

Let $x$ liters of water be added to the solution.

* New total volume of the solution $=(12+x)\text{liters}$
* New concentration of acid $=20\%$

Using the formula for the new concentration:

$$\frac{\text{Volume of Acid}}{\text{New Total Volume}}=20\%$$

$$\frac{3.6}{12+x}=\frac{20}{100}$$

$$\frac{3.6}{12+x}=\frac{1}{5}$$

Cross-multiplying to solve for $x$:

$$3.6\times 5=12+x$$

$$18=12+x$$

$$x=18-12=6\text{liters}$$

Therefore, $6\text{liters}$ of water must be added to the solution.