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mixture problem2

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buoyant Really wants to Beat The GMAT! Default Avatar
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mixture problem2 Post Sun Jan 18, 2015 12:46 pm
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  • 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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    Post 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 GMAT Titan
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    Post 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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    Post 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.

    The correct answer is D.

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    Post 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.

    Final Answer: D

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    subhakimi Just gettin' started! Default Avatar
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    Post 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

    Thanked by: buoyant

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