A mixture of milk and water contains $12\frac{1}{2}$ of water. How much water should be added to $200\text{gallons}$ of such a mixture so that the new mixture may contain $37\frac{1}{2}$ of water?

The initial amount of mixture is $200\text{gallons}$.
Initial water percentage $=12\frac{1}{2}\%=\frac{25}{2}\%$.

$$\text{Initial quantity of water}=200\times \frac{25}{200}=25\text{gallons}$$

$$\text{Initial quantity of milk}=200-25=175\text{gallons}$$

Let $x$ gallons of water be added. The amount of milk remains unchanged ($175\text{gallons}$).
In the new mixture, water is $37\frac{1}{2}\%=\frac{75}{2}\%=\frac{3}{8}$.
Therefore, milk will be $100\%-37\frac{1}{2}\%=62\frac{1}{2}\%=\frac{125}{2}\%=\frac{5}{8}$ of the new total mixture.

$$\frac{5}{8}\times (200+x)=175$$

$$200+x=\frac{175\times 8}{5}$$

$$200+x=35\times 8$$

$$200+x=280$$

$$x=80\text{gallons}$$