If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a six-percent vinegar solution, what was the concentration of the original solution?
Choose the correct choice.
A) 19.3% B) 31% C) 16.67% D) 15.5% E) 12.5%
Let X be the quantity of non-vinegar in the strong vinegar solution   Thus vinegar quantity will be 12 - X   When 50 ounces of water were added the percentage of vinegar becomes 6%, thus (12 - X)/62 = 6/100   From this equation X = 8.28   Answer (12-8.28)/12 = 31%   ANSWER:31%
<result>B</result>
If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a six-percent vinegar solution, what was the concentration of the original solution?