Thirty pounds of a salt-water solution contains fifteen percent salt. how much water must be added to weaken the solution to ten percent salt? a 1.5 lb b 3 lb c 15 lb

C. 15 lb


The 30 pounds of 15% solution contains 30*.15 = 4.5 lb of salt. In order for that amount of salt to comprise a 10% solution, 4.5/0.10 = 45 lb of solution are needed. 45 -30 = 15 lb of water must be added.

(1/x + 1)^2 + 6 (1/x + 1) +5