A right circular cylindrical vessel contains a mixture of milk and water in the ratio $4:5$. When $x$ ml of this mixture is removed and replaced by $x$ ml of pure milk, the ratio of milk to water in the resulting mixture becomes $57:60$. The amount of mixture removed is $108$ ml less than the initial volume of the mixture in the vessel.

Given that the radius of the cylindrical vessel is $0.7 ext{ cm}$ and $37 ext{ cm}^3$ of the vessel remains empty, find the height of the vessel.

Question

A right circular cylindrical vessel contains a mixture of milk and water in the ratio $4:5$. When $x$ ml of this mixture is removed and replaced by $x$ ml of pure milk, the ratio of milk to water in the resulting mixture becomes $57:60$. The amount of mixture removed is $108$ ml less than the initial volume of the mixture in the vessel.

Given that the radius of the cylindrical vessel is $0.7 	ext{ cm}$ and $37 	ext{ cm}^3$ of the vessel remains empty, find the height of the vessel.

Accepted Answer

**Solution:**

1. Let the initial quantity of the mixture in the vessel be $V$ ml.

2. The initial ratio of milk to water is $4:5$, so:
   - Milk = $\frac{4}{9}V$
   - Water = $\frac{5}{9}V$

3. When $x$ ml of mixture is removed and replaced by $x$ ml of pure milk, the new quantities become:
   - Milk after removal: $\frac{4}{9}V - \frac{4}{9}x$ (since the removed mixture has milk in the same ratio)
   - Water after removal: $\frac{5}{9}V - \frac{5}{9}x$

4. After adding $x$ ml of pure milk, milk quantity becomes:
   $$\left(\frac{4}{9}V - \frac{4}{9}xight) + x = \frac{4}{9}V + \frac{5}{9}x$$

5. Water remains:
   $$\frac{5}{9}V - \frac{5}{9}x$$

6. The new ratio of milk to water is given as $57:60$, so:
   $$\frac{\frac{4}{9}V + \frac{5}{9}x}{\frac{5}{9}V - \frac{5}{9}x} = \frac{57}{60}$$

7. Simplify the ratio:
   $$\frac{4V + 5x}{5V - 5x} = \frac{57}{60}$$

8. Cross-multiplied:
   $$60(4V + 5x) = 57(5V - 5x)$$
   $$240V + 300x = 285V - 285x$$

9. Rearranged:
   $$240V - 285V = -285x - 300x$$
   $$-45V = -585x$$
   $$45V = 585x$$
   $$V = 13x$$

10. Given that the quantity of mixture sold is $108$ ml less than the initial quantity:
    $$V - x = 108$$

11. Substitute $V = 13x$:
    $$13x - x = 108$$
    $$12x = 108$$
    $$x = 9$$

12. Calculate $V$:
    $$V = 13 	imes 9 = 117 	ext{ ml}$$

13. Volume of mixture in the vessel is $117 	ext{ ml} = 117 	ext{ cm}^3$ (since $1 	ext{ ml} = 1 	ext{ cm}^3$).

14. Total volume of the vessel = volume of mixture + empty volume = $117 + 37 = 154 	ext{ cm}^3$

15. Volume of cylinder formula:
    $$V = \pi r^2 h$$

16. Substitute $V = 154$, $r = 0.7$ cm:
    $$154 = \pi 	imes (0.7)^2 	imes h$$
    $$154 = \pi 	imes 0.49 	imes h$$

17. Calculate $h$:
    $$h = \frac{154}{\pi 	imes 0.49} = \frac{154}{1.54} \approx 100 	ext{ cm}$$

**Answer:** $100 	ext{ cm}$

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