Question

A vessel contain's mixture of $24l$ milk and $6l$ water and a second vessel contains a mixture of $15l$ milk and $10l$ water. How much mixture of milk and water should be taken from the first and the second vessel separately and kept in a third vessel so that the third vessel may contain a mixture of $25l$ molk and $10l$ water?

Answer 1

Let $x$ $l$ of mixture be taken from 1st vessel and $y$ $l$ of the mixture be taken from 2nd vessel and kept in 3rd vessel so that $(x+y)$ $l$ of the mixture may contain a mixture of $25$ $l$ of milk and $10$ $l$ of water.
A mixture of $x$ $l$ form 1st vessel contains $\frac{24}{30}x=\frac{4}{5}x$ $l$ of milk and $\frac{x}{5}$ $l$ of water and a mixture of $y$ $l$ from 2nd vessel contains $\frac{3y}{5}$ $l$ of milk and $\frac{2y}{5}$ $l$ of water.
$\therefore$ $\frac{4}{5}x+\frac{3}{5}y=\mathrm{25.........}\left(i\right)$
$\frac{x}{5}+\frac{2}{5}y=\mathrm{10........}\left(ii\right)$
Solving equation $\left(i\right)$ and $\left(ii\right)$ $x=20$ litres and $y=15$ litres.