$\frac{3}{5}$ of a mixture that has $\%40$ of alcohol is poured. If we put water as amount of poured mixture into the mixture, what would be the alcohol ratio?
If there's $\%40$ of alcohol, then we need to have $\%60$ of water. However, I couldn't keep this on with given info. Below is a table which I tried to draw.
\begin{array}{l c c c} \text{mixture} & \text{water (l)} & \text{alcohol (l)} & \text{total (l)}\\ \hline \text{solution 1} & 60 & 40 & 20\\ \text{solution 2} & 20 & 0 & 20\\ \text{solution 3} & 80 & 0 & 80\\ \hline \text{new mixture} & 160 & 40 & 200 \end{array}
Where solution 2 represents to the poured mixture. Hence we get $160 \cdot \frac{100}{100} = \boxed{\%16}$
Regards!