kprabhala.mba wrote:The following method is called alligation. It's an easy way to handle weighted average problems.
The proportion needed of each starting fraction is the positive difference between the other 2 fractions.
Mitch, just curious how the alligation technique would work for more than 2 fractions.
e.g. 3 alloys with 3 materials in different proportions are mixed to form a compound alloy.
The equation method would definitely be tedious with multiple unknowns. I hope such questions
would not appear in GMAT as they are computationally intensive (and time-consuming)
I wouldn't use alligation for a mixture that contains more than 2 elements. Most mixture problems can be solved easily by plugging in our own values or by plugging in the answer choices.
Here is a GMATPrep problem in which 3 different elements are combined. If you don't want to be exposed to the problem and its solution, please read no further.
Three grades of milk are 1%, 2%, 3% fat by volume. If x gallons of the 1% grade, y gallons of the 2% grade, and z gallons of the 3% grade are mixed to give x+y+z gallons of a 1.5 % grade, what is x in terms of y and z?
1) y+3z
2) (y+z)/4
3) 2y + 3z
4) 3y+z
5) 3y+4.5z
If we use 0 gallons of z, then equal amounts of x (1%) and y (2%) will yield a mixture that is 1.5% fat.
Plug in x=2, y=2, and z=0.
The question asks for the value of x=2. This is our target.
Now we plug y=2 and z=0 into all the answer choices to see which yields our target of 2.
Only answer choice A works:
y + 3z = 2 + 3*0 = 2.
The correct answer is
A.
Here's an OG problem that can be solved easily by plugging in the answers:
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 %
Answer choice C: Resulting mixture is 40% X
Let X=40, Y=60.
Ryegrass in X = .4*40 = 16.
Ryegrass in Y = .25*60 = 15.
Total ryegrass = 16+15 = 31.
Total ryegrass/Total mixture = 31/100 = 31%.
Since the resulting percentage of ryegrass must be just a bit smaller (30%), we need to use just a bit less of X (since it contains a smaller percentage of ryegrass than does Y).
The correct answer is
B.
The problem above also could be solved by using alligation:
Proportion needed of X = 30-25 = 5.
Proportion needed of Y = 40-30 = 10.
Percentage of X in final mixture = 5/15 = 33.33%.
Hope this helps!
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