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If 12 ounces of a strong vinegar solution are diluted with [#permalink]
18 Jul 2010, 23:07

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68% (02:58) correct
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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?

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%

Let the concentration of the original solution be \(x%\).

\(3%\) of vinegar in \(50+12=62\) ounces of vinegar solution came from \(12\) onces of \(x%\) strong vinegar solution (as in 50 ounces of water there is no vinegar at all), so: \(12*x=0.03(12+50)\) --> \(x=0.155=15.5%\).

Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
27 Sep 2012, 12:40

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Let X be the quantity of non-vinegar in the strong vvinegar solution Thus vinegar quantity will be 12 - X When 50 ounces of water were added the percentage of vinegar becomes 3%, thus (12 - X)/62 = 3/100 From this equation X = 10.14 Answer (12-10.14)/12 = 15.5%

Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
13 May 2013, 22:16

Vinegar Quantity = 12 ounces(x% of concentration) = percentage of vinegar in water vinegar mixture i.e, 12*x/100 = 3*(50+12)/100 x=15.5 % Original vinegar concentration

Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
20 Sep 2013, 13:28

bibha 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?

A. 19.3% B. 17% C. 16.67% D. 15.5% E. 12.5%

This is a type of question where homogenous solution(Water) is mixed with a mixture Therefore concentration is inversely proportional to volume i.e Concentation(final) X Volume(Final) = Concentration(initial) X volume(initial) 0.3 X 62 = Concentration(initial) X 12 Concentration (initial) = 15.5% Hence D

Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
21 Sep 2013, 02:15

Expert's post

adityapagadala wrote:

I am not sure where i going wrong.

Let the total volume is v ounces, hence we need to find out (12/v)*100. When 50 ounces of water is added the total volume is increased to V+50

Given (12/v+50) * 100 = 3 on solving we get v=350. hence 12/350 * 100 = 3.4

Can some one tell me whats wrong in this way?

What has the volume has to do with the problem?

50 ounces of 0% solution (water) is added to 12 ounces of x% solution, resulting in 62 ounces of 3% solution. The question asks to find the value of x.

Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
23 Mar 2015, 05:53

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