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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]
13 Feb 2004, 10:02
0% (00:00) correct
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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 ?
I believe kpadma needs to be kicked ( Just kidding ) for not showing his work.
Here is my anlysis
Let us say they are both mixed in equal proprtions
X contributes 100 gms
Y contributes 100 gms
X+Y = 200 gms
Out X we get 40 gms of R and out of Y we get 25 gms of R
so total = 65 gms
now 65/(200) = 32.5% Since the stem says 30% we can confidently say that they are not mixed in equal proprtions
Let us say Y is added two times making it contribute
50 gms of R and 150 gms of F
then total R = 50+40 = 90
and X+Y = 100+200 = 300
Now net R / total weight = 90/300 = 30% which is what the stem says
So X/(X+Y) = 100(100+200) = 33.3%
Can some one reverse engineer this and show it by set of equations ?