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shaftbmf
 
 

Official Guide #240

by shaftbmf Thu Sep 13, 2007 2:05 am

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
Guest
 
 

response...

by Guest Thu Sep 13, 2007 12:33 pm

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.
StaceyKoprince
ManhattanGMAT Staff
 
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Location: Montreal
 

by StaceyKoprince Thu Sep 13, 2007 5:17 pm

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
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