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

Entered answer:

✅ Correct Answer: 45

We have a container filled with milk. We perform two identical operations:

Draw out 9 litres and replace with water

Draw out 9 litres again and replace with water

After these operations, the ratio of milk to water becomes 16:9.

Let's say the container has capacity T litres.

When we draw out 9 litres from a container of capacity T, we're removing $\dfrac{9}{T}$ fraction of the total mixture.

The remaining fraction after each operation is:

$\text{Remaining fraction} = 1 - \dfrac{9}{T} = \dfrac{T-9}{T}$

After first operation:

Milk concentration = $\dfrac{T-9}{T}$ (since initially it was pure milk)

After second operation:

Milk concentration = $\dfrac{T-9}{T} \times \dfrac{T-9}{T} = \left(\dfrac{T-9}{T}\right)^2$

The final ratio is milk:water = 16:9

This means: $\dfrac{\text{milk}}{\text{total volume}} = \dfrac{16}{16+9} = \dfrac{16}{25}$

Therefore: $\left(\dfrac{T-9}{T}\right)^2 = \dfrac{16}{25}$

$\dfrac{T-9}{T} = \dfrac{4}{5}$

$5(T-9) = 4T$

$5T - 45 = 4T$

$T = 45$

Notice that $\dfrac{16}{25} = \left(\dfrac{4}{5}\right)^2$

This tells us that $\dfrac{4}{5}$ of the mixture remains each time, meaning $\dfrac{1}{5}$ is drawn out each time.

Since $\dfrac{1}{5} \times T = 9$ litres are drawn out:

$T = 9 \times 5 = 45$ litres

Therefore, the capacity of the container is 45 litres.

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