The ratio of wine to water in vessel A is $5:3$ and in vessel B is $8:5$. When the contents of both vessels are combined, the total quantity of wine is $80$ liters and the total quantity of water is $49$ liters. What is the total quantity of the mixture in vessel B?

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**Solution:**

1. Let the quantity of mixture in vessel A be $x$ liters and in vessel B be $y$ liters.

2. The ratio of wine to water in vessel A is $5:3$. So,
   - Wine in A = $\frac{5}{8}x$
   - Water in A = $\frac{3}{8}x$

3. The ratio of wine to water in vessel B is $8:5$. So,
   - Wine in B = $\frac{8}{13}y$
   - Water in B = $\frac{5}{13}y$

4. Total wine combined = $80$ liters:
   $$\frac{5}{8}x + \frac{8}{13}y = 80$$

5. Total water combined = $49$ liters:
   $$\frac{3}{8}x + \frac{5}{13}y = 49$$

6. Multiply the first equation by 104 (LCM of 8 and 13) to clear denominators:
   $$104 	imes \left(\frac{5}{8}x + \frac{8}{13}yight) = 104 	imes 80$$
   $$13 	imes 5x + 8 	imes 8y = 8320$$
   $$65x + 64y = 8320 \quad \quad (1)$$

7. Multiply the second equation by 104:
   $$104 	imes \left(\frac{3}{8}x + \frac{5}{13}yight) = 104 	imes 49$$
   $$13 	imes 3x + 8 	imes 5y = 5096$$
   $$39x + 40y = 5096 \quad \quad (2)$$

8. Multiply equation (2) by 5 and equation (1) by 4 to eliminate $y$:
   - Equation (2) $	imes 5$: $195x + 200y = 25480$
   - Equation (1) $	imes 4$: $260x + 256y = 33280$

9. Subtract the first from the second:
   $$260x + 256y - (195x + 200y) = 33280 - 25480$$
   $$65x + 56y = 6800$$

(Note: This step needs correction; let's re-check elimination carefully.)

Actually, to eliminate $y$, multiply (1) by 5 and (2) by -4:

- (1) $	imes 5$: $325x + 320y = 41600$
   - (2) $	imes (-4)$: $-156x - 160y = -20384$

Adding:
   $$325x - 156x + 320y - 160y = 41600 - 20384$$
   $$169x + 160y = 21216$$

This still has $y$. Let's try eliminating $x$ instead.

10. Multiply (1) by 39 and (2) by -65:
    - (1) $	imes 39$: $2535x + 2496y = 324480$
    - (2) $	imes (-65)$: $-2535x - 2600y = -331240$

Adding:
    $$0x - 104y = -6760$$
    $$-104y = -6760$$
    $$y = \frac{6760}{104} = 65$$

11. Substitute $y=65$ into equation (1):
    $$65x + 64 	imes 65 = 8320$$
    $$65x + 4160 = 8320$$
    $$65x = 8320 - 4160 = 4160$$
    $$x = \frac{4160}{65} = 64$$

12. The question asks for the total quantity of mixture in vessel B, which is $y = 65$ liters.

**Answer:** $65 	ext{ liters}$

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