A $12\text{litre}$ solution of acid and water contains $30\%$ acid. How much water (in litres) must be added to get a solution having $20\%$ acid?

Let's track the quantity of pure acid in the solution since adding water does not alter the amount of acid.
Step 1: Find the initial quantity of acid.

$$\text{Initial volume of solution}=12\text{litres}$$

$$\text{Acid content}=30\%\text{ of }12=\frac{30}{100}\times 12=3.6\text{litres}$$

Step 2: Formulate the new solution mixture.
Let $x$ litres of water be added to the mixture.

$$\text{New total volume of solution}=(12+x)\text{litres}$$

In the final solution, the acid concentration drops to $20\%$, but its absolute quantity remains $3.6\text{litres}$:

$$20\%\text{ of }(12+x)=3.6$$

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

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

**Step 3: Solve for $x$.**

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

$$12+x=18$$

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

Therefore, $6\text{litres}$ of water must be added.