A solution has acid to water ratio of 6:4. If 5 L of water is added, the new ratio becomes 6:5. What is the original quantity of acid in the solution?

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**Solution:**

1. Let the original quantities of acid and water be $6x$ and $4x$ liters respectively.

2. After adding $5$ liters of water, the quantity of acid remains $6x$, and water becomes $4x + 5$.

3. The new ratio is given as $6:5$, so:
   $$\frac{6x}{4x + 5} = \frac{6}{5}$$

4. Cross-multiply:
   $$5 	imes 6x = 6 	imes (4x + 5)$$
   $$30x = 24x + 30$$

5. Simplify:
   $$30x - 24x = 30$$
   $$6x = 30$$
   $$x = 5$$

6. Original acid quantity:
   $$6x = 6 	imes 5 = 30 	ext{ L}$$

**Answer:** $30 	ext{ L}$

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