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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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26 Feb 2008, 19:46

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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?

Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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18 Aug 2009, 00:00

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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 %

METHOD 1: Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g. Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.

So we can now equate the parts of the ryegrass in the mixture as:

223. 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 (33.33%) (C) 40% (D) 50% (E) 66 2/3 (66.66%)

Shortest way please.

Just focus on ryegrass. X has 40% ryegrass and Y has 25% ryegrass to give 30% ryegrass in mixture. Therefore, X:Y = 1:2 (because distance of X and Y from average is in the ratio 10:5 i.e. 2:1. If this in unclear, check out http://gmatclub.com/forum/tough-ds-105651.html#p828579) Hence X is 33.33% in the mixture. _________________

Re: Seed mixture X is ryegrass and bluegrass [#permalink]

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24 Sep 2010, 23:41

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IMHO this task is better to solve with absolute numbers. Say, there are 100 liters of X (40+60) and Y (25+75) that add to 200 liters in total. The weight of reygrass in the united mixture is 0,3*200=60. The weight of reygrass in X is 40. 40/60=2/3=33,33%

240. 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%

B. x = weight of x 40 x + 25 (1-x) = 30 solve for x, x = 33.33% _________________

Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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09 Mar 2013, 00:25

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Expert's post

Sachin9 wrote:

Is this Question correct?

It is asking. what percent of the weight of this mixture is X ?

And we are answering percent of the ryegrass in X /ryegrass in mixture..

Yes it is. The question tells us that mixture X and mixture Y (both of which are themselves mixtures of ryegrass and bluegrass) are mixed together to give a super mixture. We need to find the weight of mixture X in this super mixture.

We know the % of ryegrass in X and in Y and we also know the % of ryegrass in the super mixture (i.e. the average % of ryegrass when you mix X and Y). This tells us the ratio of X and Y in the super mixture i.e. in what ratio were X and Y mixed together (using our standard weighted averages method). When you have the ratio of X and Y, you can say what the % of mixture X is in the super mixture.

Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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21 Jul 2013, 18:43

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1. x and y should have been mixed in the ratio, 30-25 / 40-30 = 1/2 to get the resulting mixture 2. So the weight of x in the resulting mixture is 1/3 * 100 % = 33.33 % _________________

Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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11 Nov 2013, 21:01

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sondenso wrote:

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%

Its very simple ..

30/100(x+y)= 40/100(x) + 25/100(y) after solving we'll get, y=2x ... and we know x + y = 100(Total mixture)

Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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29 Apr 2016, 09:38

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sondenso wrote:

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 great problem because we can solve it using equations and substitution, or we can employ some estimation because the equation we create is a weighted average. Let’s first do the equation-substitution method.

Equation-Substitution Method

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 or 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, and we have:

40x + 25y = 30x + 30y

10x = 5y

2x = y

The question asks what percent 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:

This is a method I would suggest using only if you are short on time.

We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the overall weight of the mixture is 30% ryegrass.

Since 30% is closer to 25% than it is to 40%, we know that there is more of mixture Y than there is of mixture X, but not by a large amount. So let’s analyze our answer choices.

A) 10%

This tells us that the mixture is made up of 10% x and 90% y. That discrepancy is much too large to be correct, based on the overall weight of the mixture.

B) 33 1/3%

This tells us that the mixture is made up of 33 1/3% x and 66 2/3% y. This could be the answer.

C) 40%

This tells us that the mixture is made up of 40% x and 60% y. This could be the answer.

D) 50%

This tells us that the mixture is made up of 50% x and 50% y. Since we know that y is weighted more heavily in the mixture, we know this cannot be the answer.

E) 66.67%

This tells us that the mixture is made up of 66 2/3% x and 33 1/3% y. Since we know that y is weighted more heavily in the mixture, we know this cannot be the answer.

So now we have our answer choices narrowed down to B and C. However, there is something interesting about answer choice B. Notice that if we were to add together answer choice B (33 1/3%) and answer choice E (66 2/3%), our sum would be 100%. These are strategic answer choices, because the GMAT is hoping that if we make a mistake in our calculations we will determine that x represents 66 2/3% of the mixture rather than 33 1/3% of the mixture. Since answer choice C (40%) does not have a corresponding trap answer there is a higher likelihood that the answer is B (33 1/3%) than it is C (40%).

As mentioned at the beginning, the surest solution technique is to create an equation to solve it and actually complete the math. However, if you are short on time or have trouble setting up the equation, estimation is a method you can employ.

The answer is B. _________________

Jeffrey Miller Jeffrey Miller Head of GMAT Instruction

Re: Seed mixture X is ryegrass and bluegrass [#permalink]

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13 Oct 2010, 02:11

Sorry, while I understand the explanation, I am bit perplexed on where I am making the mistake

RyeGrass Others Mixture A 40lb 60lb 100lb Mixture B 25lb 75lb 100lb ============================ 65lb 135lb 200lb

Here 200lbs of mixture will have 65lb of RyeGrass.

In other words, 100lbs of mixture will have 32.5lbs of Ryegrass.

The question is saying - " What % of the weight of the mixture is X" the mixture here is 100lb and Mixture A is 50lb, so why isn't it (50/100)*100=50%?

I know this is NOT a correct answer but what am I missing?

Re: Seed mixture X is ryegrass and bluegrass [#permalink]

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26 Jan 2011, 08:18

one more way to solve this prob.

since mixture X has 40 percent ryegrass and 60 percent bluegrass i.e they are in ratio of ( 40/60) or ( 2/3) similarly mixture y has 25 percent ryegrass and 75 percent fescue i.e they are in ratio of ( 25/75) or (1/3) where as resultant is a mixture in which 30% is ryegrass i.e it is in ratio of ( 30/70) or ( 3/7)

then ryegrass in mixture x = 2/5 ryegrass in mixture y is =1/4

ryegrass in mixture(x+y)= 3/10

then they are mixed in ratio as quantity of y/ quantity of x= ((2/5)-(3/10))/ ((3/10)-(1/4))= 2 /1

then quantity of x is 1/3 or 33.3% _________________

Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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17 Jul 2012, 11:34

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 ?

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