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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 threepercent vinegar solution, what was the concentration of the original solution? A. 19.3% B. 17% C. 16.67% D. 15.5% E. 12.5%
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19 Jul 2010, 00:19




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Re: mixture [#permalink]
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19 Jul 2010, 00:58
i will go with Option D,
New mixture = 62 ounces, which means 1.86 ounce for vinegar
Vinegar source is only from solution, so 1.86 of 12 ounce
which means 15.5 % of the solution




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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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27 Sep 2012, 13:40
Let X be the quantity of nonvinegar 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 (1210.14)/12 = 15.5%
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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27 Sep 2012, 14:46
V1*S1=V2*S2. V1=12, S1=S1; V2=(50+12)=62, S2=3%. THEN, S1= (62*3%)/12=0.155=15.5%



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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



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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 threepercent 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



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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20 Sep 2013, 19:06
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?



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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20 Sep 2013, 19:12
i am not sure where i am going wrong.
Let V be the initial volume of solution and we need to find (12/v)*100 and there by v
After adding 50 ounces of water, the concentration is 3% i.e
12/(v+50) * 100 = 3=> v= 350
so 12/350 * 100 = 3.4
Can some one tell where i am going wrong?



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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21 Sep 2013, 03:15



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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20 Oct 2013, 00:01
Solved it through allegation rule. Vinegar in Strong solution = x Vinegar in water = 0 By allegation rule 3/(x3) = 12/50 (12/50 coz, they are mixed in the same proportions) 150 = 12x36 12x = 186 x= 15.5 So the initial conc of vinegar was 15.5%
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If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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29 Jul 2015, 00:56
x/x+(12x)+50=1.86/62, where xvinegar, 12x is nonvinegar
x/62=1.86/62, x=1.86, so (1.86/12)*100=15.5%
OR
100x/62=3 => x=1.86, (1.86/12)*100=15.5%
D



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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 threepercent 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



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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18 Jul 2016, 20:54
let x=% vinegar in original solution 12x=.03(12+50) x=1.86/12=15.5%



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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20 Jul 2016, 03:57
62 ounce = 3% vinegar solution, which is 1.86 ounce of vinegar vinegar in original solution= (1.86/12)*100 = 15.5%



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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 threepercent 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%. using this method  http://www.veritasprep.com/blog/2011/04 ... mixtures/12/50 = 3/x3 x=15.5%



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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27 Feb 2018, 11:13
Hi All, This question can be set up using the Weighted Average Formula. We're told that.... 1) 12 oz of a strong solution are mixed with 2) 50 ox. of a water solution (with 0 vinegar in it) 3) The resulting mix is 3% vinegar. We're asked for the concentration of vinegar in the 12 oz. solution. X = % concentration in the 12 oz solution (12(X) + 50(0)) / (12 + 50) = .03 12X + 0 = 62(.03) 12X = 1.86 X = 1.86/12 X = 186/1200 X = .155 Thus, the original solution is 15.5% vinegar. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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09 Apr 2018, 15:32
I used the formula C1*V1 = C2*V2
C1 * 12 = (3/100) * 62
12*C1 = 186/100 C1= (1.86/12)*100 = 15.5%



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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]
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19 Apr 2018, 13:43
bibha wrote: If 12 ounces of a strong vinegar solution are diluted with 50 ounces of water to form a threepercent vinegar solution, what was the concentration of the original solution?
A. 19.3% B. 17% C. 16.67% D. 15.5% E. 12.5% When 50 ounces of water is added to the 12 ounces of original solution, we get 62 ounces of a NEW solution. The new solution is 3% vinegar. 3% of 62 = 1.86So, the NEW solution contains 1.86 ounces of pure vinegar. These 1.86 ounces of pure vinegar came from the ORIGINAL solution. The ORIGINAL solution had a volume of 12 ounces. 1.86 ounces was pure vinegar. 1.86/ 12 = what percent? Well, 1.8/12 = 15%, so 1.86/ 12 must be a little bit more than 15%. Answer: D Cheers, Brent
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