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OG Mixtures Problem

This topic has 3 expert replies and 1 member reply
aaggar7 Master | Next Rank: 500 Posts Default Avatar
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OG Mixtures Problem

Post Mon Mar 11, 2013 7:53 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Seed mixture X is 40 percent ryegrass and 60 percent
    bluegrass by weight; seed mixture Y is 25 percent
    ryegrass and 75 percent fescue. If a mixture of X and
    Y contains 30 percent ryegrass, what percent of the
    weight of the mixture is X ?
    (A) 10%
    (B) 33 1/3%
    (C) 40%
    (D) 50%
    (E) 66 2/3%

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    Anju@Gurome GMAT Instructor
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    Post Mon Mar 11, 2013 9:06 pm
    aaggar7 wrote:
    Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?
    This can be treated as an weighted average problem where 30 is the weighted average of 25 and 40 and we need to find out the weight of 40 in the average.

    Say, the weight of 40 is x.
    Hence, weight of 25 is (1 - x)

    So, 40x + 25(1 - x) = 30
    --> (40x - 25x) = (30 - 25)
    --> 15x = 5
    --> x = 1/3

    Required percentage = 100/3 = 33" %

    The correct answer is B.

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    Post Mon Mar 11, 2013 9:16 pm
    Here is another different approach based on the same concept.

    In any weighted average, thee weight of any end point in the average will be (Difference between the other end-point and the weighted average)/(Difference between the two end-points)

    Hence, here weight of 40 in the average = |25 - 30|/|40 - 25| = 5/15 = 1/3

    Required percentage = (1/3)*100 = 33"

    The correct answer is B.[/quote]

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    Thanked by: aaggar7
    aaggar7 Master | Next Rank: 500 Posts Default Avatar
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    Post Tue Mar 12, 2013 12:26 am
    Thanks Anju Smile

    Post Tue Mar 12, 2013 2:36 am
    For a visual solution, check the second half of my post here:

    http://www.beatthegmat.com/ratios-fraction-problem-t115365.html

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