Two vessels $A$ and $B$ contain solutions of acid and water. In $A$ and $B$, the ratios of acid and water are $7:3$ and $4:1$, respectively. They are mixed in the ratio $1:2$. What is the ratio of acid and water in the resulting solution?

Let's normalize the total volumes of vessels $A$ and $B$:In $A$, Acid $:$ Water $=7:3$ (Total $=10$)In $B$, Acid $:$ Water $=4:1$ (Total $=5$)To make the totals equal, multiply the ratio of $B$ by 2:In $B$, Acid $:$ Water $=8:2$ (Total $=10$)Now they are mixed in the ratio $1:2$.Multiply vessel $A$ 's components by 1 and vessel $B$ 's components by 2:From $A$: Acid $=7\times 1=7$, Water $=3\times 1=3$ From $B$: Acid $=8\times 2=16$, Water $=2\times 2=4$ Total Acid in the mixture $=7+16=23$ Total Water in the mixture $=3+4=7$ Thus, the final ratio of acid to water is $23:7$.