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by Stuart@KaplanGMAT » Fri Sep 18, 2009 8:14 pm
uptowngirl92 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?

19.3%
17%
16.67%
15.5%
12.5%
We know that the new solution is 3% vinegar and has a total of 62 ounces of liquid.

Using the percent equation:

% = part/whole

3/100 = ounces vinegar/62

(3/100)62 = ounces of vinegar

Now, before we actually calculate, let's think about what we're doing next.

We want to know the % vinegar in the original solution. So, once again:

% = part/whole * 100%

% = (3/100)(62)/12 * 100%

% = (186)*(1/12)%

Doing some cancellation:

% = (93/6)%

% = 15.5%... choose D
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by Vena » Fri Nov 11, 2011 1:08 am
The resulted solution contains: 0.03(12+50)=1.86 ounce of vinegar.
Adding 50 ounce of water does not change the amount of vinegar in the original solution and the final one.
Hence, the %vinegar in the original solution is: (1.86/12)*100=15.5%
Choose D
Could anyone tell me if my approach is appropriate? Tnx a lot
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by shankar.ashwin » Fri Nov 11, 2011 1:12 am
Perfect :)
Vena wrote:The resulted solution contains: 0.03(12+50)=1.86 ounce of vinegar.
Adding 50 ounce of water does not change the amount of vinegar in the original solution and the final one.
Hence, the %vinegar in the original solution is: (1.86/12)*100=15.5%
Choose D
Could anyone tell me if my approach is appropriate? Tnx a lot
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by GMATGuruNY » Fri Nov 11, 2011 3:21 am
uptowngirl92 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?

19.3%
17%
16.67%
15.5%
12.5%
To make the math easy, treat the given percentages as averages.

Amount of vinegar in final solution = (total ounces)(average) = 62*3 = 186.
Amount of vinegar in added water = 0.
Average amount of vinegar in original solution = (total vinegar)/(ounces) = 186/12 = 15.5.

The correct answer is D.
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