OG Seed mixture

[gmatblood]

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%

[rijul007]

let us say the mixture has 100gm of X and ygm of Y

40 + 25% of y = 30% of (100+y)
40 + y/4 = 30 + 3y/10
3y/10 - y/4 = 10
y/20 = 10
y = 200

percentage weight of X = (100/(100+200))*100 = 33.33%

Option B

[pemdas]

let X and Y be corresponding weights, then (0.4X+0.25Y)/(X+Y)=0.3, find X-?
the previous statement can be rewritten as 40X+25Y=30X+30Y and 10X=5Y or X/Y=1/2
now X/(X+Y) or ratio of X to the total will be 1/(1+2)=1/3
this makes (1*100)/3 % or 33.33%

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%

[GMATGuruNY]

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.