Why do you need tables to solve? Don't use a more difficult method when a very easy one is right in front of you.

You know it's not D, because that would actually give you 32.5%, not 30%. It might be 40%, but it's not as easy to test as 33.33% is.

If you figure you have 100 lbs of X, then for X to be 33.33% of the end mixture, there needs to be 200 lbs of Y. So for 100 lbs of X you get 40 lbs of ryegrass, and then 25% of 200 lbs of Y = 50, so a total of 90 lbs of ryegrass and 300 lbs of mixture. 90/300 = 30%. We have a winner.

rxs0005 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 this mixture is X ?

(A) 10%

(B) 33.33 %

(C) 40%

(D) 50%

(E) 66.66 %

Can some one solve this mixture problem using Tables

thanks

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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