Question

A vessel contains mixture of $24liter$ milk and $6litre$ water and a second vessel contains a mixture of $15litre$ milk and $10litre$ water. If $M$ litre is the sum of mixture of milk and water that should be take from the first and the second vessel seperately and kept in a third vessel so that the third vessel maay contain a mixture of $25liter$ of milk and $10litre$ of water then find $\frac{M}{5}$.

Answer 1

Let $5x$ litre mixture is taken from first vessel

$\therefore$ It contains $4x$ litre milk and x litres water

Let $5y$ litre mixture is taken from second vessel

$\therefore$ It contains $3y$ litre milk and $2y$ litre water

$\therefore 4x+3y=25$ …………$\left(1\right)$

$x+2y=10$ …………..$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get

$4x+3y=25$

$-4(x+2y)=10\times 4$

$-5y=-15$

$\therefore y=3$

$\therefore x=4$

$\therefore 5x=4\times 5=20$ litre

$5y=5\times 3=15$ litre

$\therefore 5x+5y=35$ litre $=$M

$\therefore \frac{M}{5}=\frac{35}{5}=7$

$\therefore \frac{M}{5}=7$.