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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]
26 Feb 2008, 18:46
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45% (medium)
Question Stats:
70% (02:48) correct
30% (02:47) wrong based on 965 sessions
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]
17 Aug 2009, 23:00
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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:
Re: 223 Percentage [#permalink]
23 Feb 2011, 18:21
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Baten80 wrote:
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]
24 Sep 2010, 22:41
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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%
Re: 240. PS.Mixture [#permalink]
27 Feb 2008, 00:41
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sondenso wrote:
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]
08 Mar 2013, 23:25
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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]
21 Jul 2013, 17:43
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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]
11 Nov 2013, 20:01
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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 ryegrass and bluegrass [#permalink]
13 Oct 2010, 01:11
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]
26 Jan 2011, 07:18
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]
17 Jul 2012, 10:34
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 ?
Re: Mixture Problem [#permalink]
17 Jul 2012, 10:43
debabrata44 wrote:
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 explanation given as :
1=X+Y --> Y=1-X
(4/10)X+(1/4)Y = (3/10)(X+Y)
(4/10)X+(1/4)(1-X) = (3/10)(x+(1-X))
get common denominator
(16x+(10-10x))/40 = (3/10)
6x+10=12
6x=2
x=2/6 = 1/3 => 33.3%
Can anyone pls expain me how they are taking 1=X+Y.
Pls response. Thanks a lot
The question asks "what percent of the weight of this mixture is X". If we denote by X the percentage of the mixture which is of type X and by Y the percentage of the mixture which is of type Y, then together they must give 100% = 1. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
gmatclubot
Re: Mixture Problem
[#permalink]
17 Jul 2012, 10:43
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