Understanding the Initial Mixture Problem

The problem describes a cylindrical vessel containing a mixture of milk and water. We are given the initial ratio and how the mixture changes after a certain amount is sold and replaced.

The initial ratio of milk to water in the vessel is given as 5 : 3.

Finding Original Quantity and Quantity Sold (x)

Let the original quantity of the mixture in the vessel be \(Q\) ml.

Based on the initial ratio of 5 : 3, the initial quantities of milk and water are:

Initial quantity of milk = \(\frac{5}{5+3} Q = \frac{5}{8} Q\)

Initial quantity of water = \(\frac{3}{5+3} Q = \frac{3}{8} Q\)

When \(x\) ml of the total given mixture is sold, the quantities of milk and water sold are proportional to the initial ratio:

Quantity of milk sold = \(\frac{5}{8} x\)

Quantity of water sold = \(\frac{3}{8} x\)

After selling \(x\) ml, the remaining quantities in the vessel are:

Remaining milk = Initial milk - Milk sold = \(\frac{5}{8} Q - \frac{5}{8} x\)

Remaining water = Initial water - Water sold = \(\frac{3}{8} Q - \frac{3}{8} x\)

Then, \(x\) ml of pure milk is added to the vessel. The new quantities in the resultant mixture are:

Resultant quantity of milk = Remaining milk + Milk added = \(\left(\frac{5}{8} Q - \frac{5}{8} x\right) + x = \frac{5}{8} Q + \frac{3}{8} x\)

Resultant quantity of water = Remaining water + Water added = \(\left(\frac{3}{8} Q - \frac{3}{8} x\right) + 0 = \frac{3}{8} Q - \frac{3}{8} x\)

The problem states that the ratio between the resultant quantity of milk to that of water in the resultant mixture is 75 : 13.

So, we can write the equation:

\(\frac{\text{Resultant quantity of milk}}{\text{Resultant quantity of water}} = \frac{75}{13}\)

\(\frac{\frac{5}{8} Q + \frac{3}{8} x}{\frac{3}{8} Q - \frac{3}{8} x} = \frac{75}{13}\)

Multiply the numerator and denominator by 8 to simplify:

\(\frac{5Q + 3x}{3Q - 3x} = \frac{75}{13}\)

Cross-multiplying gives:

\(13(5Q + 3x) = 75(3Q - 3x)\)

\(65Q + 39x = 225Q - 225x\)

Collect like terms:

\(39x + 225x = 225Q - 65Q\)

\(264x = 160Q\)

Divide both sides by 8:

\(33x = 20Q\)

We are also given that the quantity of mixture which is sold earlier (\(x\)) was 104 ml less than the original quantity of the given mixture (\(Q\)) in the vessel:

\(x = Q - 104\)

This can be rewritten as \(Q = x + 104\).

Substitute the expression for \(Q\) into the equation \(33x = 20Q\):

\(33x = 20(x + 104)\)

\(33x = 20x + 2080\)

Subtract \(20x\) from both sides:

\(33x - 20x = 2080\)

\(13x = 2080\)

Solve for \(x\):

\(x = \frac{2080}{13}\)

\(x = 160\) ml

Now we can find the original quantity \(Q\) using \(Q = x + 104\):

\(Q = 160 + 104 = 264\) ml

So, the original quantity of the mixture in the vessel was 264 ml, and the quantity sold/\(x\)) was 160 ml.

Using the original quantity and initial ratio, we find the initial quantities of milk and water:

Initial milk = \(\frac{5}{8} \times 264 = 5 \times 33 = 165\) ml

Initial water = \(\frac{3}{8} \times 264 = 3 \times 33 = 99\) ml

Calculating Final Quantity in a New Scenario

Now, we address the specific question: If 0.25x ml of the initial mixture is sold and 0.75x ml pure milk and 0.05x ml water is added, find the final total quantity.

Initial mixture quantity = 264 ml

Initial milk = 165 ml

Initial water = 99 ml

Value of \(x\) = 160 ml.

First, 0.25x ml of the initial mixture is sold.

Quantity of mixture sold = \(0.25x = 0.25 \times 160 = 40\) ml

This 40 ml sold mixture has the same milk to water ratio (5:3) as the initial mixture.

Quantity of milk sold from this 40 ml = \(\frac{5}{8} \times 40 = 5 \times 5 = 25\) ml

Quantity of water sold from this 40 ml = \(\frac{3}{8} \times 40 = 3 \times 5 = 15\) ml

Quantities remaining in the vessel after selling 40 ml:

Remaining milk = Initial milk - Milk sold = 165 ml - 25 ml = 140 ml

Remaining water = Initial water - Water sold = 99 ml - 15 ml = 84 ml

Total remaining quantity = 140 ml + 84 ml = 224 ml (which is 264 - 40).

Next, 0.75x ml of pure milk and 0.05x ml of water are added to the vessel.

Quantity of pure milk added = \(0.75x = 0.75 \times 160 = 120\) ml

Quantity of water added = \(0.05x = 0.05 \times 160 = 8\) ml

The final quantities of milk and water in the resultant mixture are calculated by adding the quantities added to the remaining quantities:

Final quantity of milk = Remaining milk + Milk added = 140 ml + 120 ml = 260 ml

Final quantity of water = Remaining water + Water added = 84 ml + 8 ml = 92 ml

The final total quantity of the resultant mixture in the given vessel is the sum of the final quantities of milk and water.

Final total quantity = Final quantity of milk + Final quantity of water

Final total quantity = 260 ml + 92 ml = 352 ml

Final Answer

The final total quantity of the resultant mixture in the given vessel is 352 ml.