A mixture containing 4% saline solution is to be mixed with 2 quarts of a mixture which is 10% saline to obtain a solution which is 8% saline. How much of the 4% solution must be used?
Btw: No need to give me step by step. I only need the answer.
+36916 amount of saline in ingredients = saline in final soln
x = 4% .04x
10% = 2 quarts = .10(2)
Final solution = .08 (2 + x)
.04x + .1(2) = .08 ( 2+x) solve for x quarts
