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mww7786
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Post subject: OG - DS - #240 Posted: Tue May 08, 2007 9:22 am |
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I hope this message finds you well. This problem was a little difficult to set up the MGMAT way. Will you please throw some light on this.
#240
Seed mix X is 40% ryegrass and 60% bluegrass by weight. Seed mix Y is 25% ryegrass and 75% Fercue. if a mixture of X & Y contains 30% Ryegrass, what percent of the weight of the mixture is X?
a.10%
b. 33.3%
c. 40%
d. 50%
e. 66 & 2/3
answer: b
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StaceyKoprince
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Post subject: OG11 Problem Solving #240 Posted: Tue May 08, 2007 5:18 pm |
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| ManhattanGMAT Staff |
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Posts: 6077 Location: San Francisco
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I assumed this is from OG11; if not, please correct me so others can find it if need be.
Another weighted average problem.
So X contains 40% rye and 60% blue (and I'm sneezing as I think about that :)) and Y contains 25% rye and 75% Fescue.
Then I mix some unknown amount of each together and get 30% rye. Remember that X had 40% rye and Y had 25% rye - so, from the start, I can tell there's more Y than X because the percentage of rye in the final mix is closer to Y's starting point than X's. I can cross off answers D and E.
Here's the interesting thing with weighted average problems - I can actually use the proportional amount that the new mix is closer to Y (or further from X) to find the answer. The difference between 25 (%rye in Y) and 40 (%rye in X) is 15. 30% is 5 away from Y's starting point, or 1/3 of 15. 30% is 10 away from X's starting point, or 2/3 of 15. Swap those figures, and that's the percentage that X and Y contributed to the final mix! So, Y is 2/3 (or 66.6%) and X is 1/3 (or 33.3%).
_________________ Stacey Koprince Instructor Director of Online Community ManhattanGMAT
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Carla
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Post subject: Question on Weighted Averages Posted: Sat May 19, 2007 10:35 am |
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Hi Stacey
I did not see a section in the red books specifically on the topic of weighted average - if it is in there and I missed it can you let me know? I wonder if you have a specific formula or method to use when facing a weighted average problem...
Thanks!
Carla
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Jeff
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Post subject: Weighted Average Posted: Sat May 19, 2007 7:24 pm |
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Carla -
Here's one approach: Let p = percentage of Seed X in the mixture. Then you know that (1-p) = percentage of Seed Y in the mixture, since there are only two seed types and their total must sum to 100%.
Therefore 40p + (1-p) * 25 = 30
15 p = 5
p = 1/3 = 33 1/3%
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StaceyKoprince
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Post subject: Posted: Mon May 21, 2007 2:04 am |
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| ManhattanGMAT Staff |
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Posts: 6077 Location: San Francisco
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Nice technique, Jeff. Also, Carla, the approach I described in my answer to the question is the most common approach I use.
_________________ Stacey Koprince Instructor Director of Online Community ManhattanGMAT
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