Problem 1. In two tubs, there are $4 \frac{1}{6}$ and $3 \frac{1}{2}$ liters of water respectively. How much water needs to be transferred from the first tub to the second tub so that after the transfer, both tubs have the same amount of water? Explain your answer!
Solution. In the first barrel, there are $\frac{25}{6}$ liters, while in the second barrel, there are $\frac{7}{2}$ liters. From the condition of the problem, we have $\frac{25}{6}-x=\frac{7}{2}+x$. We can rewrite the last equation as $2 x=\frac{25}{6}-\frac{7}{2}$, or $x=\frac{1}{2} \cdot \frac{4}{6}=\frac{2}{6}=\frac{1}{3}$. Therefore, we need to transfer $\frac{1}{3}$ liters from the first barrel to the second barrel to have the same amount of water in both barrels.