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| OG - DS - #240 |
| OG11 Problem Solving #240 |
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Stacey Koprince
MGMAT STAFF
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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%). |
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| Question on Weighted Averages |
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Carla
Guest
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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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| Weighted Average |
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Jeff
Guest
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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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Stacey Koprince
MGMAT STAFF
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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.
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| OG - DS - #240 |
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