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555-605 (Medium)|   Mixture Problems|                                          
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 22 May 2024, 04:11
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Weighted average method

Assumption: X+Y=100
AVG=30 (30% X+Y)
Focusing only on Ryegrass --> A2=40 (40% X) A1= 25 (25% Y) --> w2=X & w1=Y

w1/w2= A2-AVG/AVG-A1 --> 40-30/30-25= 10/5 --> 10w1 + 5w2=15 --> 5/15*100 (100 is X+Y) = 100/3 = X *100% = 100/3 % --> 33,3%­
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 05 Aug 2024, 07:55
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­"Hello. I'd like to buy the book that has exercises like the ones above. Could you please tell me the title of this book? Thank you."
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TheGMATStrategy
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 06 Aug 2024, 17:45
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Anhminh
­"Hello. I'd like to buy the book that has exercises like the ones above. Could you please tell me the title of this book? Thank you."
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­It's called the "Official Guide for GMAT® Review". There's typically a new version every year. You can find the current one here: https://www.mba.com/exam-prep/gmat-offi ... stion-bank

Does that help?
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 13 Sep 2025, 05:46
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Will try to Use some reasoning...25% - Y and X - 40%.....
Means there is more of Y than of X.....So DE go out ....A seems too less for X ...That also goes out...Now B and C...
If X is 33.33% that means Y is 66% ...means double of X...Lets see..25% + 25% (Since its twice in value) + 40%..Gives us 90%/3....30% Bingo.....average ...
B is the answer
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 03 Oct 2025, 22:35
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Let's see the solution by Allegations:
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 25 Feb 2026, 15:35
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sondenso
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. \(33\frac{1}{3}\)%

C. 40%

D. 50%

E. \(66\frac{2}{3}\)%
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B




Nick Slavkovich, GMAT/GRE tutor with 20+ years of experience

[email protected]
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 31 Mar 2026, 08:14
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Hi Jeff, Can we solve this problem also with the matrix method?
JeffTargetTestPrep


We are given information about ryegrass, bluegrass, and fescue. However, we are told the mixture of X and Y is 30 percent ryegrass. Thus, we only care about the ryegrass, so we can ignore fescue and bluegrass.

We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the weight of the combined mixture is 30% ryegrass. With x representing the total weight of mixture X, and y representing the total weight of mixture Y, we can create the following equation:

0.4x + 0.25y = 0.3(x + y)

We can multiply the entire equation by 100:

40x + 25y = 30x + 30y

10x = 5y

2x = y

The question asks what percentage of the weight of the mixture is x.

We can create an expression for this:

x/(x+y) * 100 = ?

Since we know that 2x = y, we can substitute in 2x for y in our expression. So, we have:

x/(x+2x) * 100 = x/(3x) * 100 = 1/3 * 100 = 33.33%

Answer: B
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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig 25 Aug 2026, 17:53
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sondenso
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. \(33\frac{1}{3}\)%

C. 40%

D. 50%

E. \(66\frac{2}{3}\)%
Show more
This official question provides a great example of the balance point shortcut for weighted averages.

Mixture X is 40% ryegrass and mixture Y is 25% ryegrass. The combined mixture is 30% ryegrass, closer to Y at 25% than to X at 40%. Thus there must be more of mixture Y.

How much more? For the combined mixture to balance at 30%, the ratio of the quantities must equal the ratio of the distances from 30%. From 40 to 30 is 10, and from 25 to 30 is 5, so the ratio of the quantities is:
\(\frac{10}{5} = \frac{2}{1}\)

Since there is more of mixture Y, you have 2 parts Y to 1 part X. Thus \(X = \frac{1}{3} = 33 \frac{1}{3}\)%

I explain in more detail here:

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