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If 12 ounces of a strong vinegar solution are diluted with
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bibha
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If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  18 Jul 2010, 23:07
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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%
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Bunuel
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Re: mixture [#permalink] New post  18 Jul 2010, 23:19
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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: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  19 Apr 2018, 12:43
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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?

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 = 1.86
So, 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

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Re: mixture [#permalink] New post  18 Jul 2010, 23:58
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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] New post  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%

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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  27 Sep 2012, 13:46
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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] New post  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
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  20 Sep 2013, 13:28
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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?

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] New post  20 Sep 2013, 18:06
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] New post  20 Sep 2013, 18:12
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] New post  21 Sep 2013, 02:15
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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] New post  19 Oct 2013, 23:01
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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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If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  28 Jul 2015, 23:56
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%

D
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  02 May 2016, 09:20
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
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  18 Jul 2016, 19:54
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] New post  20 Jul 2016, 02:57
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] New post  25 Aug 2017, 21:14
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] New post  27 Feb 2018, 10:13
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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.

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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  09 Apr 2018, 14:32
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] New post  09 Aug 2019, 21:41
12x = 0.03(62)
or
12x = 3/100 * 62
1200x = 186
600x = 93
x= 93/600 *100/1 %
x= 93/6 %
x= 15.5%
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