A mixture is composed of $8$ parts of brandy and $3$ parts of water. After adding $28$ litres of water, if the mixture contains brandy one half as much as water, then how many litres of brandy does it contain?
To me, initial $3x$ and $8x$
$8x = \dfrac{3x+28}{2}$ leads to solution
But in text, this is done as $8x = \dfrac{(3x+28)3}{2}$
Is my calculation right? I guess it is a simple one. But to make sure I am right, please comment
Let $x$ be the amount in L of the mixture. Then the water amount is: $\dfrac{3x}{11}$ L, and the brandy abount is $\dfrac{8x}{11}$L. After pouring in $28$ L of water, the new amount is $(x+28)$ L, and one third of this amount is brandy. So the brandy amount is: $\dfrac{x+28}{3}$ L. Since the brandy amount does not change before and after pouring of water into the mixture, it must be true that: $\dfrac{8x}{11} = \dfrac{x+28}{3} \to24x=11x+308 \to 13x= 308 \to x = \dfrac{308}{13} $ L. Thus the mount of brandy is: $\dfrac{8}{11}\cdot \dfrac{308}{13} = 17.23$ L.