A 10quart radiator contains a 40antifreeze solutionHow much

A 10-quart radiator contains a 40%antifreeze solution.How much of the solution needs to be drained out and replaced with pure antifreeze in order to raise the solution to 70% antifreeze?

Solution

Lets assume replaced antifreeze solution be x

active - active + active = active
0.40*10 - 0.40x + 1.00x = 0.70*10

4 -0.4x +x = 7

0.6x = 3

x = 3/0.6= 5 quarts

So, 100% antifreeze solution of 5 quarts should be replaced to get 70% antifreeze