Question

A radiator holds 8 quarts of fluid, 20% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 50% antifreeze?

Answer 1

After draining x quarts the amount of pure antifreeze in the remaining fluid is 20% of 8-x, that is, 0.2(8-x). The amount of pure antifreeze added is x quarts, and the amount of pure antifreeze in the new solution is 50% of 8 quarts=4 quarts. Therefore, 0.2(8-x)+x=4, 1.6-0.2x+x=4, 0.8x=2.4, so x=3 quarts. So 3 quarts is drained off.