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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 ?

The way I approached this is: assume out of the 30% ryegrass, x% is from X hence, amount of bluegrass in the mixture = 3x/2 similarly, (30-x)% is from Y hence, amount of fescue = 3(30-x)

x + 3x/2 + (30-x) + 3(30-x) =100 solve for x => 40/3 3x/2 => 20

20% of the mixture is bluegrass, therefore, total for X=40/3+20=100/3 = 33 1/3 B

Sure, R=Ryegrass, B=Bluegrass, F=Fescue X: 40% R, 60% B Y: 25% R, 75% F Mixture(lets call it M): 30% R, some B, some F Question: What% of M is X

Since M is mixture, and ryegrass is M is from both X and Y, Hence, the 30% of Ryegrass, will be the sum of ryegrass which comes from X and from Y. Assume that the % of ryegrass contributed by X is x. Total ryegrass = 30% % from X= x% Therefore, percentage from Y= (30-x)%

Now, in X the ratio of R:B = 40:60 Hence, if R=x, B=(60/40)x=3x/2

Similarly, in Y the ratio of R:F = 25:75 Hence, if R=30-x, F=(75/25)(30-x) = 3(30-x)

Adding all gives you 100%, so, x + 3x/2 + (30-x) + 3(30-x) =100 solve for x => 40/3 B= 3x/2 => 20

To find total % of X in M, you have to add amount of B and amount of R from X = 20 + 40/3 = 33.33