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?

Question: 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?

1. 19.3%
2. 17%
3. 16.67%
4. 15.5%
5. 12.5%

“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?”– is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.0

Answer
Approach Solution 1

The resulting 62 ounces of the solution have 3% of pure vinegar.

So, it has \(62*\frac{3}{100}\) ounces of pure vinegar or 1.86 oz.

If the 12 ounces of the original had 1.86 oz. of the pure vinegar, it must have been of the following concentration:

\(\frac{1.86}{12}=15.5\)

Correct option: D

Approach Solution 2

Let the concentration of the original solution be x%

3 of vinegar in 50 + 12 = 62 ounces of vinegar solution came from 12 ounces of x% strong vinegar solution (as in 50 ounces of water there is no vinegar at all), so: 12 * x = 0.03 (12 + 50)

Solving gives x = 0.155 = 15.5%

Correct option: D

Approach Solution 3

\(C_1\)= initial concentration

\(V_1\)= initial volume = 12 oz.

\(C_2\)= final concentration

\(V_2\)= final volume = (12 + 50) = 62 oz.

Formula used:

\(C_1 V_1=C_2V_2\)

\(C_1\)* 12 = 3% * 62

\(C_1\)= 15.5%

Correct option: D

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