2. In an $8 \mathrm{~kg}$ mixture, there is $65 \%$ alcohol and 35 \% water. How much water needs to be added so that the diluted mixture has 45 \% water?
Solution. Since 1% of the mixture is $80 \mathrm{gr}$, the mixture contains 2800 grams of water and 5200 grams of alcohol. Let $x$ be the amount of water, in grams, that needs to be added. Then $2800+x$ should be $45 \%$ of $8000+x$, i.e., $2800+x=0.45(8000+x)$, from which we get $x=1454.(54)$.

Therefore, to obtain a mixture with 45 \% water, 1454.54 grams of water need to be added.