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%
How do you solve? strategically guessing, i eliminated A, C,D and assumed E. thanks
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- shaftbmf
- Guest
response...
your strategic guess paid off.
you are trying to solve for X / (X + Y)
You can set up an equation to solve:
.40x + .25y = .30(x+y)
once you get the relative proportion of x to y, you can plug in to get X/(X+Y). = 33 1/3%
Also, If you understand weighted averages, you could look at the problem and realize the answer is B.
you are trying to solve for X / (X + Y)
You can set up an equation to solve:
.40x + .25y = .30(x+y)
once you get the relative proportion of x to y, you can plug in to get X/(X+Y). = 33 1/3%
Also, If you understand weighted averages, you could look at the problem and realize the answer is B.
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9360
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
Really tough. Yes, this is a weighted average problem.
In normal averages, we can just add up each item and divide by the number of items because they are all weighted equally (eg, if you have 2 things, they are both weighted at 50%). In weighted averages, different things count more - this is akin to having your tests count more than homework assignments toward your final grade in a class, for example.
So X is 40% rye and 60% blue
Y is 25% rye and 75% fescue
If X+Y rye is 30%, what percent of the mix is X (vs Y)?
The easiest way to handle this type of weighted average problem is to look at the spread between the relevant item in the two mixtures, X and Y. In X, rye is 40%. In Y, rye is 25%. When the two are mixed, rye is 30%.
Draw a number line and place these numbers on it.
25 30 35 40
Y - M - - - X (M = mixture of X+Y)
M is 5 units higher than Y and 10 units lower than X. Because M is closer to Y, this means Y is weighted more heavily - that is, there's more than 50% Y in the mix and less than 50% X in the mix. How much more? Well, going from Y to X, M splits the line into 5/15, or 1/3, and 10/15, or 2/3, portions. Those are the two proportions of mix I'm dealing with then - Y is weighted more heavily, so Y is 2/3 and X is weighted less heavily so X is 1/3. They've asked me for X, so the answer is 1/3 or 33 1/3%. B
In normal averages, we can just add up each item and divide by the number of items because they are all weighted equally (eg, if you have 2 things, they are both weighted at 50%). In weighted averages, different things count more - this is akin to having your tests count more than homework assignments toward your final grade in a class, for example.
So X is 40% rye and 60% blue
Y is 25% rye and 75% fescue
If X+Y rye is 30%, what percent of the mix is X (vs Y)?
The easiest way to handle this type of weighted average problem is to look at the spread between the relevant item in the two mixtures, X and Y. In X, rye is 40%. In Y, rye is 25%. When the two are mixed, rye is 30%.
Draw a number line and place these numbers on it.
25 30 35 40
Y - M - - - X (M = mixture of X+Y)
M is 5 units higher than Y and 10 units lower than X. Because M is closer to Y, this means Y is weighted more heavily - that is, there's more than 50% Y in the mix and less than 50% X in the mix. How much more? Well, going from Y to X, M splits the line into 5/15, or 1/3, and 10/15, or 2/3, portions. Those are the two proportions of mix I'm dealing with then - Y is weighted more heavily, so Y is 2/3 and X is weighted less heavily so X is 1/3. They've asked me for X, so the answer is 1/3 or 33 1/3%. B
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
3 posts Page 1 of 1