Solving the Milk and Water Mixture Problem

This problem involves a classic concept of mixture replacement. We start with a container of pure milk, and a certain amount is repeatedly removed and replaced with water. We need to find the ratio of water to milk after this process is performed multiple times.

Understanding the Process

Initially, the container has 50 liters of pure milk.

In each step, 10 liters of the mixture are taken out, and 10 liters of water are added. When a mixture is taken out, the components are removed in proportion to their current quantities in the mixture.

Since the container always holds 50 liters after each replacement, the total volume remains constant.

Calculating Milk Remaining After Each Step

Let $V$ be the initial volume (50 liters) and $x$ be the amount removed and replaced (10 liters).

After the first step, the amount of milk remaining is:

• Initial milk = 50 liters
• Amount of milk removed = $\text{Fraction removed} \times \text{Amount of milk} = \frac{x}{V} \times 50 = \frac{10}{50} \times 50 = 10$ liters
• Milk remaining after 1st step = $50 - 10 = 40$ liters
• Water added = 10 liters
• Total volume = $40 + 10 = 50$ liters (mixture of 40L milk + 10L water)

After the second step, 10 liters of this mixture are removed. The ratio of milk to water in the mixture is currently 40:10 or 4:1. So, in the 10 liters removed:

• Amount of milk removed = $\frac{4}{4+1} \times 10 = \frac{4}{5} \times 10 = 8$ liters
• Amount of water removed = $\frac{1}{4+1} \times 10 = \frac{1}{5} \times 10 = 2$ liters
• Milk remaining after removing 10L of mixture = $40 - 8 = 32$ liters
• Water remaining after removing 10L of mixture = $10 - 2 = 8$ liters
• 10 liters of fresh water are added.
• Milk remaining after 2nd step = $32$ liters
• Water remaining after 2nd step = $8 + 10 = 18$ liters
• Total volume = $32 + 18 = 50$ liters

Using a Formula for Repeated Replacement

A more efficient way to calculate the amount of the original substance (milk) remaining after $n$ repetitions of this process is using the formula:

$\text{Amount of milk remaining} = \text{Initial amount of milk} \times \left(1 - \frac{\text{Amount replaced}}{\text{Total volume}}\right)^{\text{Number of repetitions}}$

Let $M_n$ be the amount of milk remaining after $n$ repetitions, $M_0$ be the initial amount of milk, $x$ be the amount replaced each time, and $V$ be the total volume.

$M_n = M_0 \left(1 - \frac{x}{V}\right)^n$

In this problem:

• $M_0 = 50$ liters
• $x = 10$ liters
• $V = 50$ liters
• $n = 3$ (the process is repeated thrice)

Substitute these values into the formula:

$M_3 = 50 \left(1 - \frac{10}{50}\right)^3$

$M_3 = 50 \left(1 - \frac{1}{5}\right)^3$

$M_3 = 50 \left(\frac{5 - 1}{5}\right)^3$

$M_3 = 50 \left(\frac{4}{5}\right)^3$

$M_3 = 50 \times \left(\frac{4^3}{5^3}\right)$

$M_3 = 50 \times \left(\frac{64}{125}\right)$

Now, simplify the expression:

$M_3 = \frac{50 \times 64}{125}$

$M_3 = \frac{(2 \times 25) \times 64}{5 \times 25}$

Cancel out the common factor of 25:

$M_3 = \frac{2 \times 64}{5}$

$M_3 = \frac{128}{5}$ liters

Calculating the Amount of Water

The total volume of the mixture remains 50 liters. The mixture consists of milk and water. So, the amount of water is the total volume minus the amount of milk remaining.

Amount of water = Total volume - Amount of milk remaining

Amount of water = $50 - \frac{128}{5}$

To subtract, find a common denominator:

Amount of water = $\frac{50 \times 5}{5} - \frac{128}{5}$

Amount of water = $\frac{250}{5} - \frac{128}{5}$

Amount of water = $\frac{250 - 128}{5}$

Amount of water = $\frac{122}{5}$ liters

Finding the Final Ratio of Water to Milk

The question asks for the ratio of water and milk finally (Water : Milk).

Ratio = Amount of water : Amount of milk

Ratio = $\frac{122}{5} : \frac{128}{5}$

Since both amounts have the same denominator (5), we can write the ratio as:

Ratio = $122 : 128$}

Simplifying the Ratio

To simplify the ratio $122 : 128$, we find the greatest common divisor (GCD) of 122 and 128. Both numbers are even, so they are divisible by 2.

• $122 \div 2 = 61$
• $128 \div 2 = 64$

So the simplified ratio is $61 : 64$. The numbers 61 and 64 have no common factors other than 1, so this is the simplest form.

Therefore, the final ratio of water to milk is $61 : 64$.

Conclusion

After repeating the process of removing 10 liters of mixture and replacing it with 10 liters of water thrice, starting with 50 liters of pure milk, the final ratio of water and milk is $61 : 64$.

Revision Table: Mixture Replacement Concepts

Additional Information: Mixture Problems and Ratios

Mixture problems are common in quantitative aptitude and competitive exams. They often involve calculating the proportions or amounts of different components when substances are mixed or replaced.

• Understanding Ratios: A ratio expresses the relative size of two or more values. A ratio $a:b$ means that for every $a$ units of the first quantity, there are $b$ units of the second quantity.
• Concentration: The concentration of a substance in a mixture is the amount of the substance divided by the total volume of the mixture (often expressed as a percentage or fraction). In this problem, the concentration of milk decreases with each step.
• Constant Volume vs. Changing Volume: Some mixture problems involve mixing substances without replacement, where the total volume changes. Other problems, like this one, maintain a constant total volume through the replacement process. The formula used here applies specifically to constant volume replacement processes.
• Applications: Mixture problems have practical applications in chemistry (calculating solution concentrations), finance (calculating average interest rates), and various industrial processes.

Solving mixture problems often requires careful step-by-step calculation or the application of specific formulas derived for these scenarios. Understanding how the composition changes when a part of the mixture is removed and replaced is key.