Suppose 8 pints of a 15 alcohol solution is mixed with 2 pin

Suppose 8 pints of a 15% alcohol solution is mixed with 2 pints of a 90% alcohol solution. What is the concentration of alcohol in the new 10-pint mixture? The concentration of alcohol in the new 10-pint mixture is %. (Type an integer or a decimal.)

Solution

Let x = the concentration of alcohol in the new mixture.

Then,

8(0.15) + 2 (0.9) = 10 x

1.2 + 1.8 = 10x

3 = 10x

x = 3/10 = 0.3

So the required concentration = 0.3 *100 = 30%