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Seed mixture X is 40 percent ryegrass and 60 percent

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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 07 Oct 2005, 10:55
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 1/3%
(C) 40%
(D) 50%
(E) 66 2/3 %

Need explanation pls

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SS
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 [#permalink] New post 07 Oct 2005, 11:15
let x be amount k and y be amount 1-k in the resulting mixture

k*40/100 + (1-k)*25/100 = 30/100

so K = 1/3 or 33 1/3%
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 [#permalink] New post 07 Oct 2005, 11:36
The equation I get is
40% of ryegrass in X + 25% ryegrass in Y = 30% ryegrass in X+Y
i.e. 0.40X + 0.25Y = 0.30(X+Y)
=> 0.40X - 0.30X = 0.30Y - 0.25Y
=> 0.1X = 0.05Y
=> X/Y = 1/2, we can now work with this ratio to get
X/X+Y = 1/1+2 = 1/3 = 33 1/3% or B.
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 [#permalink] New post 07 Oct 2005, 12:02
well on this type of question you can also logically guess...

we know that Mix X has 40% Ryegrass

we know the Mix Y has 25% Ryegrass.

now if we both X+Y are combined the resulting ryegrass is 30%

well we logically see that since resulting rye grass is less than what it was in Mix X, then Mix X has to be less than 50% ...however we can also see that the resulting MIX has has 30% Ryegrass which is more than Mix Y, then answer choice 10% can also be easily elminated...we are now left with 2 choices 33% or 40% (you are down to 50-50 chance of getting it right)


anywar here is how I would solve it

40X + (1-X)25=30

15X=5

x=1/3 or 33.33%
  [#permalink] 07 Oct 2005, 12:02
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