Bunuel
If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a three-percent vinegar solution, what was the concentration of the original solution?
A. 19.3%
B. 17%
C. 16.67%
D. 15.5%
E. 12.5%
Another approach: There is actually no need to get too specific here, provided you have a decent grasp of how to work with percentages. Yes, as has been pointed out already, you will need to figure out a way to derive 3 percent of (12 + 50) ounces, but you can do this using mental math quite easily.
1) 10 percent of any number is that number with the decimal pushed one place to the left (e.g., 10 percent of 62 is 6.2)
2) 5 percent of any number is half the value of ten percent of that number (e.g., 5 percent of 62 is 6.2/2, or 3.1)
3) 1 percent of any number is that number with the decimal pushed two places to the left (e.g., 1 percent of 62 is 0.62)
Since we know we are dealing with 3 percent, we can just triple the value for 1 percent:
\(3(0.62)=1.86\)
We know there
must be 1.86 oz. of vinegar in the solution. Since water contains no vinegar, these 1.86 oz. have to come from the original 12 oz. of vinegar. Just work your way to the target value of 1.86 oz.
1) 10 percent of 12 oz. is 1.2 oz. (too low, but more than halfway to the target number)
2) 5 percent of 12 oz. is 1.2/2, or 0.6 oz.
\(1.2+0.6=1.8\)
We have added up 15 percent of the 12-oz. solution, and we need just six one-hundredths of an ounce more. But rather than derive the exact value, we can simply test what would happen if we added 1 percent to what we have:
3) 1 percent of 12 oz. is 0.12 oz.
\(1.8+0.12>1.86\)
The answer cannot be 16 percent or greater, and since we know we need more than a 15-percent concentration of vinegar in the original solution,
the answer must be (D). Not that I necessarily encourage it, but I completed this problem entirely in my head. (It took about a minute, and I am guessing that many aspirants are just as good at math(s) as I.)
Good luck with your studies.
- Andrew