Aradiator contains 10 quarts of fluid, 30% of which is antifreeze. how much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze?

Let x be the volume of fluid removed and the volume of pure antifreeze that is added. The concentration of antifreeze in the fluid is 0.3, the concentration in pure antifreeze is 1 and that in the final solution is 0.4 The volume of the final solution is 10.
(10 - x)(0.3) + x = 10(0.4)
0.3 + 0.7x = 0.4
x = 1/7 quarts

The volume that should be drained is 1/7 quarts

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess