Question

A container whose capacity is $60$60 litre contains milk and water in the ratio $3:2$. How much quantity of the mixture should be replaced with pure milk, so that in the final mixture, ratio of milk to water is $7:3$?

Answer 1

In 60 litre of mixture,

The quantity of milk $=\frac{3}{5}\times 60$

$=36\text{litre}$

So, the quantity of water will be $=60-36$

$=24\text{litre}$

Let $x$ liters of mixture is replaced.

So, remaining milk after replacement $=36-\frac{3}{5}x+x$

$=36+\frac{2x}{5}$

Now, the ratio of milk in comparison with total amount of mixture shoulbe be $7:10$ as per the question,

$\frac{36+\frac{2x}{5}}{60}=\frac{7}{10}$

$\Rightarrow 36+\frac{2x}{5}=42$

$\Rightarrow \frac{2x}{5}=42-36$

$\Rightarrow \frac{2x}{5}=6$

$\therefore x=15\text{litre}$