If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a eight-percent vinegar solution, what was the concentration of the original solution?	Choices
A) 9.3% B) 17.3% C) 26.67% D) 35.5% E) 41.3%
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 8%, thus (12 - X)/62 = 8/100   From this equation X = 7.04   Answer (12-7.04)/12 = 41.3%   ANSW41.3%R:41.3%
<result>E</result>
If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a eight-percent vinegar solution, what was the concentration of the original solution?