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Vinegar solution

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by das.ashmita » Tue Jul 24, 2012 1:33 am
Hi
I am quite week in Mixture problems.

I got this question from Papgust's compilation @https://www.beatthegmat.com/download-gma ... 59366.html. Can someone please help we with a clear approach.

31. 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?
19.3%
17%
16.67%
15.5%
12.5%

OA: D
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by Anurag@Gurome » Tue Jul 24, 2012 1:44 am
das.ashmita wrote: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?
Say, the concentration of the original solution was x%.
Hence, in 12 ounce solution there was 12x/100 ounce vinegar.

In the final (12 + 50) = 62 ounce solution there is 12x/100 ounce vinegar.

So, (12x/100)/62 = 3/100
--> 12x = 3*62
--> x = 3*62/12 = 31/2 = 15.5

The correct answer is D.
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by GMATGuruNY » Tue Jul 24, 2012 3:28 am
das.ashmita wrote:Hi
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?
19.3%
17%
16.67%
15.5%
12.5%
Treat the percentages as if they were AVERAGES.

Total amount of vinegar in 62 ounces of the 3% mixture = 62*3 = 186.
Amount of vinegar in 50 ounces of water = 0.
Thus, the average amount of vinegar in 12 ounces of the original solution = 186/12 = 15.5.

The correct answer is D.
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by NicoleWhite » Thu Jul 26, 2012 7:59 am
I think the easiest way is to create a formula similar to how the OG solves mixture problems. So, we're combining 12 ounces of a solution with concentration x and 50 ounces of water with concentration 0 to get 62 ounces of a solution with concentration 3:

12x + 50(0) = 62*3
12x = 186
x = 15.5
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