A jar contains a mixture of two liquids A and B in the ratio $4:1$. When $10$ liters of this mixture is removed and then $10$ liters of liquid B is added to the jar, the new ratio of A to B becomes $2:3$. What was the initial quantity of liquid A in the jar?

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**Solution:**

1. Let the initial total quantity of the mixture be $x$ liters.

2. The initial ratio of liquids A to B is $4:1$, so:
   - Quantity of A = $\frac{4}{5}x$
   - Quantity of B = $\frac{1}{5}x$

3. When $10$ liters of the mixture is removed, it is removed in the same ratio $4:1$:
   - Amount of A removed = $10 	imes \frac{4}{5} = 8$ liters
   - Amount of B removed = $10 	imes \frac{1}{5} = 2$ liters

4. After removal, the quantities left are:
   - A left = $\frac{4}{5}x - 8$
   - B left = $\frac{1}{5}x - 2$

5. Then, $10$ liters of liquid B is added, so new quantity of B becomes:
   $$\left(\frac{1}{5}x - 2ight) + 10 = \frac{1}{5}x + 8$$

6. The new ratio of A to B is $2:3$, so:
   $$\frac{\frac{4}{5}x - 8}{\frac{1}{5}x + 8} = \frac{2}{3}$$

7. Cross-multiplied:
   $$3\left(\frac{4}{5}x - 8ight) = 2\left(\frac{1}{5}x + 8ight)$$

8. Simplify:
   $$3 	imes \frac{4}{5}x - 24 = 2 	imes \frac{1}{5}x + 16$$
   $$\frac{12}{5}x - 24 = \frac{2}{5}x + 16$$

9. Bring like terms together:
   $$\frac{12}{5}x - \frac{2}{5}x = 16 + 24$$
   $$\frac{10}{5}x = 40$$
   $$2x = 40$$
   $$x = 20$$

10. Initial quantity of A:
    $$\frac{4}{5} 	imes 20 = 16 	ext{ liters}$$

**Answer:** $16 	ext{ Lt.}$

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