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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 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 %

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X: 40% ryegrass and Y: 25% ryegrass, since the mixture of X and Y has 30% ryegrass we know that it has more Y than X, but let's work out the exact breakdown:
Let x = the % of Y and 1-x = the % of X
25x + 40(1-x) = 30
25x + 40 - 40x = 30
10 = 15x
2/3 = x
% of X is 1-x, which is 1/3 so the answer is B. 33.33%
answered Jun 5 by Beginner (1 point)
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Let M = x + y
M = New mixture
x = Mixture X
y = Mixture Y
What do we need to find ?? => (x/M)*100
Equating Ryegrass in the mixture -
.4x + 0.25y = 0.3M
.4x + 0.25(M-x) = 0.3M
.4x + 0.25M - 0.25x = 0.3M
.15x = .05M
x/M = 1/3
answered Feb 17, 2016 by Guru (5,628 points)
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