Suppose 8 pints of a 26% alcohol solution is mixed with 2 pints of an 80% alcohol solution. What is the concentration of alcohol in the new 10-pint mixture?

Step 1 ：Let the volume of the 26% alcohol solution be \(8\) pints and the volume of the 80% alcohol solution be \(2\) pints.

Step 2 ：The total volume of the mixture is \(8 + 2 = 10\) pints.

Step 3 ：The amount of alcohol in the 26% solution is \(8 \times 0.26 = 2.08\) pints.

Step 4 ：The amount of alcohol in the 80% solution is \(2 \times 0.80 = 1.6\) pints.

Step 5 ：The total amount of alcohol in the mixture is \(2.08 + 1.6 = 3.68\) pints.

Step 6 ：The concentration of alcohol in the new mixture is \(\frac{3.68}{10} = 0.368\) or \(36.8\%\).

Step 7 ：\(\boxed{36.8\%}\) is the concentration of alcohol in the new 10-pint mixture.