A container holds two liquids, A and B, mixed in the ratio $8 : 5$. When $13$ liters of this mixture is removed and replaced entirely with liquid B, the new ratio of A to B in the container becomes $1 : 1$. What was the initial quantity of liquid A in the container?

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**Solution:**

1. Let the total volume of the mixture in the container be $x$ liters.

2. Initially, the quantities of liquids A and B are in the ratio $8 : 5$, so:
   - Liquid A = $\frac{8}{13}x$
   - Liquid B = $\frac{5}{13}x$

3. When $13$ liters of mixture is removed, the amount of A and B removed will be proportional to their initial ratio:
   - A removed = $13 	imes \frac{8}{13} = 8$ liters
   - B removed = $13 	imes \frac{5}{13} = 5$ liters

4. After removal, the remaining quantities are:
   - A remaining = $\frac{8}{13}x - 8$
   - B remaining = $\frac{5}{13}x - 5$

5. The $13$ liters removed is replaced entirely with liquid B, so new quantity of B is:
   $$\left(\frac{5}{13}x - 5ight) + 13 = \frac{5}{13}x + 8$$

6. After replacement, the ratio of A to B becomes $1 : 1$, so:
   $$\frac{\frac{8}{13}x - 8}{\frac{5}{13}x + 8} = 1$$

7. Cross-multiplied:
   $$\frac{8}{13}x - 8 = \frac{5}{13}x + 8$$

8. Simplify:
   $$\frac{8}{13}x - \frac{5}{13}x = 8 + 8$$
   $$\frac{3}{13}x = 16$$

9. Multiply both sides by $\frac{13}{3}$:
   $$x = 16 	imes \frac{13}{3} = \frac{208}{3}$$ liters

10. The initial quantity of liquid A is:
    $$\frac{8}{13}x = \frac{8}{13} 	imes \frac{208}{3} = \frac{8 	imes 208}{13 	imes 3} = \frac{1664}{39} = \frac{128}{3} 	ext{ liters}$$

**Answer:** $\frac{128}{3}$ liters

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