gmatblood 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 the mixture is X ?
(A) 10%
(B) 33.1/3%
(C) 40%
(D) 50%
(E) 66.2/3%
Approach 1: ALLIGATION
X = 40% ryegrass
Y = 25% ryegrass
Mixture = 30% ryegrass.
Using alligation:
The proportion needed of each ingredient in the mixture is equal to the distance between the OTHER 2 percentages.
Proportion needed of X = |Percentage in mixture - percentage in Y| = |30-25| = 5.
Proportion needed of Y = |Percentage in mixture - percentage in X| = |30-40| = 10.
X:Y = 5:10 = 1:2.
Since the ratio is 1 part X for every 2 parts Y, the percentage of X in the mixture = 1/3 = 33.33%.
The correct answer is
B.
Approach 2: PLUG IN THE ANSWERS
Since the percentage of ryegrass in the mixture (30%) is closer to the percentage of ryegrass in Y (25%), more than 50% of the mixture must be Y, implying that less than 50% of the mixture must be X.
Eliminate D and E.
Answer choice C: mixture = 40% X + 60% Y
Let X = 40 liters and Y = 60 liters.
Amount of ryegrass in X = .4*40 = 16.
Amount of ryegrass in Y = .25*60 = 15.
Total rye/Total mixture = (16+15)/100 = 31%.
Since the mixture must be 30% ryegrass, a little less X is needed.
The correct answer is
B.
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