One solution contains 2 parts salt to 8 parts water, and another contains 3 parts salt to 5 parts water. How much of each should be mixed together in order to obtain 280 quarts of a solution that is 3 parts salt to 7 parts water?

Given, Solution A contains 2 parts salt to 8 parts water Solution B contains 3 parts salt to 5 parts water Target Mixture contains3 parts salt to 7 parts water:
Expressing the salt concentration in decimal form
Solution A: 2/(2 + 8) = 2/10 = 0.20 salt Solution B: 3/(3 + 5) = 3/8 = 0.375 salt Mixed Solution: s/(3 + 7) = 3/10 = 0.30 salt

  Since target mixture is 280 quarts
If the amount of 0.375 salt present in mixed solution = x Then, the amount of 0.20 salt, y = 280 – x

Using a typical mixture equation
0.375x + 0.20(280-x) = 0.30(280)
0.375x + 56 - 0.20x = 84   Subtract 56 from both sides of the equation
0.375x - 0.20x + 56 – 56 = 84 – 56 0.375x - 0.20x = 28
0.175x = 28   Divide both sides of the equation by 0.175 0.175x/0.175 = 28/0.175
x = 160   y = 280 – x y = 280 – 160 y = 120

Therefore, x = 160 quarts of the solution that contains 3 parts salt to 5 parts water y = 120 quarts of the solution that contains 2 parts salt to 8 parts water