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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  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%
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Bunuel
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If 12 ounces of a strong vinegar solution are diluted with [#permalink] 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's assume that the original solution has a concentration of \(x\%\) vinegar.

The amount of vinegar in the original solution is therefore, \(12*\frac{x}{100}\) ounces.

The amount of vinegar in \(50+12=62\) ounces of the final solution is \(\frac{3}{100}*62\).

Since only water was added to the original solution, we can equate the amount of vinegar in the final solution to that in the original solution:

\(12*\frac{x}{100}=\frac{3}{100}*62\)

\(12x=3*62\)

\(x=15.5\)

Therefore, the original solution had a concentration of 15.5% vinegar.


Answer: D
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  19 Apr 2018, 13:43
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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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Brent
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smartmundu
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Re: mixture [#permalink] 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] 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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janix
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] New post  27 Sep 2012, 14:46
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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  20 Sep 2013, 14:28
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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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adityapagadala
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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink] 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] New post  21 Sep 2013, 03:15
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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  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] New post  27 Feb 2018, 11:13
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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  05 Apr 2023, 09:58
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