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If 12 ounces of a strong vinegar solution are diluted with

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If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 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?

A. 19.3%
B. 17%
C. 16.67%
D. 15.5%
E. 12.5%
[Reveal] Spoiler: OA

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Re: mixture [#permalink]

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New post 19 Jul 2010, 00:19
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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?

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: \(12*x=0.03(12+50)\) --> \(x=0.155=15.5%\).

Answer: D.
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Re: mixture [#permalink]

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New post 19 Jul 2010, 00:58
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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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New post 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%

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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 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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New post 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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New post 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

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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 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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New post 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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New post 21 Sep 2013, 03:15
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.

Check the solution here: if-12-ounces-of-a-strong-vinegar-solution-are-diluted-with-97494.html#p751157

Hope it helps.
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 20 Oct 2013, 00:01
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Solved it through allegation rule.
Vinegar in Strong solution = x
Vinegar in water = 0

By allegation rule

3/(x-3) = 12/50 (12/50 coz, they are mixed in the same proportions)
150 = 12x-36
12x = 186
x= 15.5

So the initial conc of vinegar was 15.5%
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 29 Jul 2015, 00:56
x/x+(12-x)+50=1.86/62, where x-vinegar, 12-x is non-vinegar

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%

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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 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%
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 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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New post 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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New post 31 Jul 2017, 18:39
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]

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New post 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%.

using this method - http://www.veritasprep.com/blog/2011/04 ... -mixtures/

12/50 = 3/x-3

x=15.5%

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Re: If 12 ounces of a strong vinegar solution are diluted with   [#permalink] 25 Aug 2017, 22:14
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