OG - DS - #240

[mww7786]

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

[StaceyKoprince]

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%).

[Carla]

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

[Jeff]

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%

[StaceyKoprince]

Nice technique, Jeff. Also, Carla, the approach I described in my answer to the question is the most common approach I use.