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Mixture problem : with allegation method

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gmatquant25 Rising GMAT Star Default Avatar
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Mixture problem : with allegation method Post Wed Mar 13, 2013 10:33 am
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  • Lap #[LAPCOUNT] ([LAPTIME])
    One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

    34%
    24%
    22%
    18%
    8.5%


    I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
    many thanks !

    I started with 10% and x% in the first row

    16% in the second row

    and x-16% and 6% in the last one

    this gives me x-16/6= ? "stuck here "

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    Anju@Gurome GMAT Instructor
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    Post Wed Mar 13, 2013 10:39 am
    gmatquant25 wrote:
    One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.
    If we view this as an weighted average problem then weighted average of 10 and x is 16, where weight of 10 is 3/4 and weight of is 1/4.

    Hence, (3/4)*10 + (1/4)*x = 16
    -----> 30 + x = 4*16 = 64
    -----> x = (64 - 30) = 34

    The correct answer is A.

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    gmatquant25 Rising GMAT Star Default Avatar
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    Post Wed Mar 13, 2013 11:06 am
    thanks - I could solve this problem in every other way. I want to do it using allegation rule , It helps me solve such problems in no time .



    Last edited by gmatquant25 on Wed Mar 13, 2013 11:51 am; edited 1 time in total

    Post Wed Mar 13, 2013 1:09 pm
    gmatquant25 wrote:
    One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

    34%
    24%
    22%
    18%
    8.5%
    Here's how to solve with alligation.
    Let O = the original solution, R = the replacement solution, and M = the mixture.

    Step 1: Plot the percentages on a number line, with the two ingredients on the ends and the mixture in the middle.
    O 10%----------M=16%----------R

    Step 2: Plot the distances between the percentages.
    (distance between O and M) : (distance between M and R) is equal to the RECIPROCAL of the ratio of O to R in the mixture.
    Since R = 1/4 of the mixture, O:R = 3:1.
    Plotting the reciprocal of this ratio on the number line, we get:
    O 10%----x-----M=16%----3x----R

    Since x is the distance between O and M:
    x = 16-10 = 6.
    Since 3x is the distance between M and R:
    R = 16 + 3x = 16 + 3*6 = 34.

    The correct answer is A.

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    vishugogo Really wants to Beat The GMAT!
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    Post Thu Mar 14, 2013 7:05 am
    [quote="GMATGuruNY"]how is R = 1/4 of the mixture

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    Post Thu Mar 14, 2013 7:08 am
    [quote="Anju@Gurome"][

    how is weight of 10 is 3/4 and weight of x is 1/4

    Post Thu Mar 14, 2013 7:48 am
    vishugogo wrote:
    how is R = 1/4 of the mixture
    From the problem: ONE-FOURTH of a solution...WAS REPLACED by a second solution.
    Thus, the replacement solution -- the value of R -- is equal to 1/4 of the resulting mixture.

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