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From a container filled with milk, $9 \; \text{litres}$ of milk are drawn and replaced with water. Next, from the same container, $9 \; \text{litres}$ 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 $\text{litres},$ is

Let the capacity of the tank be $x$ liters.

We can get the ratio of $\frac{\text{Milk}}{\text{Total}} = \frac{16}{25}$

$\Rightarrow \left(\frac{x-9}{x}\right)^{2} = \frac{16}{25}$

$\Rightarrow \frac{x-9}{x} = \sqrt{\frac{16}{25}}$

$\Rightarrow \frac{x-9}{x} = \frac{4}{5}$

$\Rightarrow 5x-45=4x$

$\Rightarrow \boxed{x=45\;\text{litres}}$

Correct Answer $:45$

$\textbf{PS:}$  Suppose, a container contains $x\text{’}$ units of a liquid from which $y\text{’}$ units are taken out and replaced by water. After $`n\text{’}$ operations quantity of pure is $x\left(1-\frac{y}{x}\right)^{n}.$

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