# Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 % fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X? GMAT Explanation, Video Solution, and More Practice!

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75% fescue. If a mixture of X and Y contains 30% 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%

This is a tough weighted average/mixture question from the GMAT Official Guide that comes up a bunch in GMAT tutoring session.

## Define the Question

what percent of the weight of the mixture is X?

We know the proportion of rye grass in each mixture, X and Y, and we know the proportion of rye grass adding x and y together. What we want to infer is the actual weight of seed mixture x.

## Setup 1

There are two ways I’d approach this. The first way is faster and my preference.

25 (Y)———-30———————–40 (X)

So, just looking at the above, any idea which mixture X or Y is in greater abundance? The weighted average, 30, is closer to 25 (Y) so there must be more of Y. That means the mixture is less than 50% X so you can eliminate D and E. Based on the layout you could also eliminate A in that 10% is too extreme.

Helpful but that’s not the actual method.

25 (Y)———-30———————–40 (X)

To solve with certainty take the difference of each part with the total. So 30 – 25 and 40 – 30. 5 and 10. Add them. 15. The 15 is the denominator. Now take the differences 5 and 10 and pop them in as numerators. 5/15 and 10/15. So x = 1/3 and y = 2/3. That’s it. B.

## Setup 2

The alt. way is using work/rate T’s. Weighted average/mixture questions are the same as average rate questions so you can solve them in the same way.

## Additional GMAT Weighted Average/Mixture Practice Questions

Here’s another tough weighted average/mixture question from the GMAT Official Guide: Jackie has two solutions that are 2 percent sulfuric acid and 12 percent sulfuric acid by volume, respectively. If these solutions are mixed in appropriate quantities to produce 60 liters of a solution that is 5 percent sulfuric acid, approximately how many liters of the 2 percent solution will be required?