Question

From a container filled with milk, 9 litres of milk are drawn and replaced with water.Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

Answer 1

Correct Answer: 45

Milk and Water Replacement Problem 

Let's denote the initial volume of the container as \( V \) litres. The problem involves two operations of drawing and replacing milk with water. Each operation involves drawing 9 litres and replacing it with the same amount of water.

After the First Operation:

The remaining milk = \( V - 9 \) litres. Therefore, the fraction of milk left is \( \frac{V - 9}{V} \).

After replacing 9 litres of water, the total quantity is still \( V \) litres, and the quantity of water is 9 litres.

After the Second Operation:

The milk remaining from the previous 9 litres is also drained and replaced by water, so we apply the same fraction to the milk quantity left after the first operation:

The remaining milk = \( \frac{V - 9}{V} \times (V - 9) = \left( \frac{V - 9}{V} \right)^2 \times V \) litres

Given Ratio:

Given that the ratio of milk to water is now 16:9, we express this in terms of quantities:

\[ \left( \frac{(V - 9)^2}{V^2} \right) \times V = \frac{16}{25}V \]

Equating and Solving:

\[ \left( \frac{V - 9}{V} \right)^2 \times V = \frac{16}{25} \times V \]

\[ \left( \frac{V - 9}{V} \right)^2 = \frac{16}{25} \]

Taking the square root on both sides:

\[ \frac{V - 9}{V} = \frac{4}{5} \]

Solving for \( V \):

\[ V - 9 = \frac{4}{5}V \]

\[ 5(V - 9) = 4V \]

\[ 5V - 45 = 4V \Rightarrow V = 45 \]

Conclusion:

The calculated capacity of the container is 45 litres, which falls within the expected range.