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## mixture problem2

tagged by: Brent@GMATPrepNow

This topic has 2 expert replies and 3 member replies
buoyant Master | Next Rank: 500 Posts
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#### mixture problem2

Sun Jan 18, 2015 12:46 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
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%

OA: D

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Brent@GMATPrepNow GMAT Instructor
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Sun Jan 18, 2015 1:19 pm
buoyant 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%

OA: D
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%.
Take D

For more on this shortcut for converting fractions to percents, see my article: http://www.gmatprepnow.com/articles/shortcut-converting-fractions-percents

Cheers,
Brent

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Marty Murray Legendary Member
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Sun Jan 18, 2015 1:22 pm
buoyant 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%

OA: D
Total solution now is 12 + 50 = 62 ounces.

Total vinegar of new solution is 62 * .03 = 1.86 ounces. Since only water was added, this is also the total vinegar in the original solution.

The percentage of vinegar in the original solution is 1.86 ounces vinegar/12 ounces total solution = .155 = 15.5 percent.

Choose D.

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GMATGuruNY GMAT Instructor
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Sun Jan 18, 2015 1:42 pm
buoyant 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%
Alternate approach:

12 ounces of a solution that is x% vinegar are combined with 50 ounces of water that is 0% vinegar to yield a 62-ounce solution that is 3% vinegar:
12x + 50*0 = 62*3
12x = 186
x = 186/12 = 31/2 = 15.5.

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Sun Jan 18, 2015 4:44 pm
Hi buoyant,

If you choose a "math approach", you can see that the "steps" that you have to take are essentially the same; the various explanations in this thread point that out.

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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Rich

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subhakimi Newbie | Next Rank: 10 Posts
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Mon Jan 19, 2015 9:33 am
buoyant 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%

OA: D
You can use Alligation
X____________0
.
.
......3.......
.
.
12 ..........50

So x-3/3=50/12
Or simply we can do x=3 + 50x3/2=3+12.5=15.5

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