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Manager  Joined: 14 Apr 2010
Posts: 124
If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 68% (02:22) correct 32% (02:36) wrong based on 695 sessions

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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%
Math Expert V
Joined: 02 Sep 2009
Posts: 64314

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9
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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Manager  Joined: 12 Jun 2007
Posts: 97

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3
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
##### General Discussion
Intern  Joined: 11 Jul 2012
Posts: 39
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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2
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

Brother Karamazov
Intern  Joined: 13 Sep 2012
Posts: 1
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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1
V1*S1=V2*S2.
V1=12, S1=S1;
V2=(50+12)=62, S2=3%.
THEN, S1= (62*3%)/12=0.155=15.5%
Intern  Joined: 02 May 2013
Posts: 18
WE: Engineering (Aerospace and Defense)
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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
Intern  Joined: 01 Jan 2013
Posts: 47
Location: India
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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1
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
Intern  Joined: 29 Jan 2013
Posts: 28
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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?
Intern  Joined: 29 Jan 2013
Posts: 28
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 64314
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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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Manager  Joined: 23 May 2013
Posts: 88
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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2
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%
Director  G
Joined: 23 Jan 2013
Posts: 506
Schools: Cambridge'16
If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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
Director  V
Joined: 05 Mar 2015
Posts: 936
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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%

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
VP  D
Joined: 07 Dec 2014
Posts: 1253
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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let x=% vinegar in original solution
12x=.03(12+50)
x=1.86/12=15.5%
Director  G
Joined: 20 Feb 2015
Posts: 722
Concentration: Strategy, General Management
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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62 ounce = 3% vinegar solution, which is 1.86 ounce of vinegar
vinegar in original solution=
(1.86/12)*100 = 15.5%
Intern  G
Joined: 01 Jan 2016
Posts: 28
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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%

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%
EMPOWERgmat Instructor V
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Affiliations: EMPOWERgmat
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Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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.

GMAT assassins aren't born, they're made,
Rich
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Manager  B
Joined: 08 Sep 2016
Posts: 97
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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I used the formula C1*V1 = C2*V2

C1 * 12 = (3/100) * 62

12*C1 = 186/100
C1= (1.86/12)*100 = 15.5%
GMAT Club Legend  V
Joined: 11 Sep 2015
Posts: 4889
GMAT 1: 770 Q49 V46
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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2
Top Contributor
1
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%.

Cheers,
Brent
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Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 GMAT 5: 650 Q48 V31
GMAT 6: 600 Q38 V35
GMAT 7: 710 Q47 V41 GPA: 3
WE: Management Consulting (Consulting)
Re: If 12 ounces of a strong vinegar solution are diluted with  [#permalink]

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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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Here's how I went from 430 to 710, and how you can do it yourself: Re: If 12 ounces of a strong vinegar solution are diluted with   [#permalink] 09 Aug 2019, 21:41

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