A vessel is full of wine. Ten gallons are drawn off and replaced with water. The operation is repeated once more. If the ratio of wine to water after the second replacement is $49:32$, what is the capacity of the vessel?

If the vessel capacity is $X$ gallons, then after two such operations the fraction of wine left is $\left(\frac{X-10}{X}\right)^2$. Since wine : water $=49:32$, the fraction of wine in the final mixture is $\frac{49}{49+32}=\frac{49}{81}$. Thus $\left(\frac{X-10}{X}\right)^2=\frac{49}{81}$, so $\frac{X-10}{X}=\frac{7}{9}$. Hence $9X-90=7X$, giving $2X=90$ and $X=45$.