A right circular cylindrical vessel initially contains a mixture of milk and water in the ratio 4:5. When $x$ ml of this mixture is removed and replaced with $x$ ml of pure milk, the ratio of milk to water in the vessel becomes 57:60. It is also given that the quantity of mixture sold ($x$ ml) is 108 ml less than the initial total volume of the mixture in the vessel.

If instead, $3x$ ml of the mixture is sold and then $\frac{x}{3}$ ml of milk and $2x$ ml of water are added to the vessel, what is the total volume of the vessel in cubic centimeters (cm³)?

Question

A right circular cylindrical vessel initially contains a mixture of milk and water in the ratio 4:5. When $x$ ml of this mixture is removed and replaced with $x$ ml of pure milk, the ratio of milk to water in the vessel becomes 57:60. It is also given that the quantity of mixture sold ($x$ ml) is 108 ml less than the initial total volume of the mixture in the vessel.

If instead, $3x$ ml of the mixture is sold and then $\frac{x}{3}$ ml of milk and $2x$ ml of water are added to the vessel, what is the total volume of the vessel in cubic centimeters (cm³)?

Accepted Answer

**Solution:**

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

2. Given 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 sold and replaced by $x$ ml of pure milk, the milk quantity changes as follows:
   - Milk removed = $\frac{4}{9}x$
   - Milk after removal = $\frac{4}{9}V - \frac{4}{9}x = \frac{4}{9}(V - x)$
   - Milk after adding pure milk = $\frac{4}{9}(V - x) + x = \frac{4}{9}V - \frac{4}{9}x + x = \frac{4}{9}V + \frac{5}{9}x$

4. Water quantity after removal (no water added) = $\frac{5}{9}V - \frac{5}{9}x = \frac{5}{9}(V - x)$

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

6. Simplify the equation:
   $$\frac{4V + 5x}{5(V - x)} = \frac{57}{60} = \frac{19}{20}$$

7. Cross-multiplied:
   $$20(4V + 5x) = 19 	imes 5 (V - x)$$
   $$80V + 100x = 95V - 95x$$

8. Rearranged:
   $$80V - 95V = -95x - 100x$$
   $$-15V = -195x$$
   $$V = 13x$$

9. Given that $x$ ml is 108 ml less than the initial volume $V$:
   $$V - x = 108$$
   Substitute $V = 13x$:
   $$13x - x = 108$$
   $$12x = 108$$
   $$x = 9$$

10. So, the initial volume $V = 13x = 13 	imes 9 = 117$ ml.

11. Now, if $3x = 27$ ml of mixture is sold, the remaining volume is:
    $$117 - 27 = 90 	ext{ ml}$$

12. Milk and water in the remaining mixture are in the original ratio 4:5, so:
    - Milk left = $\frac{4}{9} 	imes 90 = 40$ ml
    - Water left = $\frac{5}{9} 	imes 90 = 50$ ml

13. Then $\frac{x}{3} = 3$ ml of milk and $2x = 18$ ml of water are added:
    - New milk volume = $40 + 3 = 43$ ml
    - New water volume = $50 + 18 = 68$ ml

14. Total volume now:
    $$43 + 68 = 111 	ext{ ml}$$

15. Since $1 	ext{ ml} = 1 	ext{ cm}^3$, the vessel is filled with $111 	ext{ cm}^3$ of mixture.

**Answer:** 111

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