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If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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19 Jul 2010, 00:07
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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]
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27 Sep 2012, 13:40
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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]
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13 May 2013, 23:16
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]
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20 Sep 2013, 14:28
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
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]
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23 Mar 2015, 06:53
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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02 May 2016, 10:20
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%
Initially the volume=12 ounces later we add 50ounces water to get final volume =50+12 =62 ounces. Concentration of vinegar in final solution=3% As amount of vinegar is same as before or after dilution A=C*V 12*x%=62*3% X=15.5% Ans D
Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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31 Jul 2017, 18:39
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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25 Aug 2017, 22:14
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%
Alligations would be easier - Consider the concentrations of the solutions. Water 0%, Vinegar x%, Final solution 3%.
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68% (01:49) correct
