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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]  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%
[Reveal] Spoiler: OA
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Re: mixture [#permalink]  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: $$12*x=0.03(12+50)$$ --> $$x=0.155=15.5%$$.

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Re: mixture [#permalink]  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]  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

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Re: If 12 ounces of a strong vinegar solution are diluted with [#permalink]  27 Sep 2012, 13:46
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]  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]  20 Sep 2013, 13:28
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]  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]  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]  21 Sep 2013, 02:15
Expert's post
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]  19 Oct 2013, 23:01
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]  23 Mar 2015, 05:53
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If 12 ounces of a strong vinegar solution are diluted with [#permalink]  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
If 12 ounces of a strong vinegar solution are diluted with   [#permalink] 28 Jul 2015, 23:56
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