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

Assume the capacity of the container is \( c \). We need to track the amount of milk that remains after taking out 9 liters twice from the container.

Step 1: Milk remaining after the first 9 liters are taken

The amount of milk remaining after the first 9 liters are taken is:

\[ \left( c \times \left( c - \frac{9}{c} \right) \right) \]

Step 2: Milk remaining after the second 9 liters are taken

The amount of milk remaining after the next 9 liters are taken is:

\[ \left( c \times \left( c - \frac{9}{c} \right) \times \left( c - \frac{9}{c} \right) \right) \]

Step 3: Milk and water ratio

After both extractions, the ratio of milk to water in the container is 16:9. This means the fraction of milk in the container is:

\[ \frac{16}{25} \]

Step 4: Setting up the equation for the milk remaining

We can write the equation for the milk remaining in terms of \( c \) as follows:

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

Step 5: Solving the equation

\[ \left( c - \frac{9}{c} \right)^2 = \frac{16}{25} \] Taking the square root of both sides: \[ c - \frac{9}{c} = \frac{4}{5} \]

Step 6: Solving for \( c \)

\[ c - \frac{9}{c} = 1 \quad \Rightarrow \quad c = 45 \]

Final Answer:

The capacity of the container is \( \boxed{45} \) liters.

Answer 2

Let the initial volume of milk in the container be \( V \). After drawing out and replacing milk with water \( n \) times, the remaining volume of milk is given by:

\[ F = V \left(1 - \frac{K}{V}\right)^n \]

where \( K \) is the volume of milk drawn each time.

In this problem:

• Number of times milk is drawn and replaced, \( n = 2 \)
• Milk drawn each time, \( K = 9 \) liters
• Final milk volume is \(\frac{16}{25}\) of the initial volume \( V \):

\[ F = V \left(1 - \frac{9}{V}\right)^2 = \frac{16}{25} V \]

Dividing both sides by \( V \), we get:

\[ \left(1 - \frac{9}{V}\right)^2 = \frac{16}{25} \]

Taking the square root:

\[ 1 - \frac{9}{V} = \pm \frac{4}{5} \]

Check the negative root first:

\[ 1 - \frac{9}{V} = -\frac{4}{5} \implies \frac{9}{V} = 1 + \frac{4}{5} = \frac{9}{5} \implies V = 5 \]

This is impossible since \( K = 9 \) liters is drawn each time, which exceeds the container's total volume.

Now check the positive root:

\[ 1 - \frac{9}{V} = \frac{4}{5} \implies \frac{9}{V} = 1 - \frac{4}{5} = \frac{1}{5} \implies V = 45 \]

Final answer:

The initial volume of milk in the container is \( \boxed{45} \) liters.