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# Seed mixture X is 40 percent ryegrass and 60 percent

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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]  26 Feb 2008, 18:46
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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?

A. 10%
B. 33 1/3%
C. 40%
D. 50%
E. 66 2/3%
[Reveal] Spoiler: OA

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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:

0.4X + 0.25(100-X) = 30
0.4X + 25 - 0.25X = 30
0.15X = 5
X = 5/0.15 = 500/15 = 100/3

So the weight of mixture X as a percentage of the weight of the mixture
= (weight of X/weight of mixture) * 100%
= (100/3)/100 * 100%
= 33%

METHOD 2
wt. of 1st mixture = x
therefore concentration of ryegrass in 1st mix = 0.4x

wt. of 2nd mixture = y
therefore concentration of ryegrass in 2nd mix = 0.25y

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

0.4x-0.3x = 0.3y - 0.25y

0.1x=0.05y
or
2x=y

so if weight of x = 50grams
weight of y = 100 grams

total weight of mix = 150 grams

percentage of x in 150 grams of mix is 150*x/100 = 50

x = 50*100/150
x = 100/3
x = 33.3%
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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%)

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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 28 Jan 2010 Posts: 30 Followers: 0 Kudos [?]: 21 [9] , given: 45 Re: Seed mixture X is ryegrass and bluegrass [#permalink] 28 Jan 2010, 03:47 9 This post received KUDOS Let M = x + y M = New mixture x = Mixture X y = Mixture Y What do we need to find ?? => (x/M)*100 Equating Ryegrass in the mixture - .4x + 0.25y = 0.3M .4x + 0.25(M-x) = 0.3M .4x + 0.25M - 0.25x = 0.3M .15x = .05M x/M = 1/3 Hence ans = 33.33%. Intern Joined: 26 Aug 2010 Posts: 23 Followers: 0 Kudos [?]: 14 [7] , given: 2 Re: Seed mixture X is ryegrass and bluegrass [#permalink] 24 Sep 2010, 22:41 7 This post received KUDOS 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% SVP Joined: 29 Aug 2007 Posts: 2496 Followers: 58 Kudos [?]: 563 [3] , given: 19 Re: 240. PS.Mixture [#permalink] 27 Feb 2008, 00:41 3 This post received KUDOS 1 This post was BOOKMARKED 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% _________________ Manager Joined: 13 Aug 2009 Posts: 153 WE 1: 4 years in IT Followers: 1 Kudos [?]: 77 [2] , given: 7 Re: Seed mixture X is ryegrass and bluegrass [#permalink] 28 Jan 2010, 21:57 2 This post received KUDOS Hi Tejal777 we can make the method 2 more simpler METHOD 2 wt. of 1st mixture = x therefore concentration of ryegrass in 1st mix = 0.4x wt. of 2nd mixture = y therefore concentration of ryegrass in 2nd mix = 0.25y 0.4x+0.25y = 0.3(x+y) 0.4x-0.3x = 0.3y - 0.25y 0.1x=0.05y or 2x=y [b]so if weight of x = 50grams weight of y = 100 grams total weight of mix = 150 grams percentage of x in 150 grams of mix is 150*x/100 = 50 x = 50*100/150 x = 100/3 x = 33.3%[/b]. what we need is x/x+y *100 we know x and y so solving it we get 33.33% Intern Joined: 26 Feb 2008 Posts: 8 Followers: 0 Kudos [?]: 1 [1] , given: 0 Re: 240. PS.Mixture [#permalink] 27 Feb 2008, 18:06 1 This post received KUDOS the answer is E.66 2/3% solve for 25x + 40 [1-x] = 30 so, x = 2/3 = 66 2/3% Math Forum Moderator Joined: 20 Dec 2010 Posts: 2031 Followers: 137 Kudos [?]: 1129 [1] , given: 376 Re: Seed mixture X is ryegrass and bluegrass [#permalink] 05 Feb 2011, 04:27 1 This post received KUDOS take X=100 In it: Ryegrass=40 Others = 60 take Y=y Ryegrass=0.25y Others = 0.75y Mix both; Total weight: X+Y=100+y In it; Weight of Ryegrass = 40+0.25y Given: 40+0.25y=0.3(100+y) 40+0.25y=30+0.3y 0.05y=10 y=200 So; total X+Y = 100+200=300 Percentage of X in it: (X/X+Y)*100 = (100/300)*100=100/3=33.33% Ans: B _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5548 Location: Pune, India Followers: 1372 Kudos [?]: 6981 [1] , given: 178 Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] 08 Mar 2013, 23:25 1 This post received KUDOS 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. For more on such questions, check out the second example in my post: http://www.veritasprep.com/blog/2011/04 ... -mixtures/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]  21 Jul 2013, 13:21
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........,,,,,,...X...,.. Y
Ryegrass:.... 40.... 0,25X
Bluegrass:... 60....
fescue:. ............. 0,75X
----------------------------------------------
Total:........ 100..... X--...---100+X

40 + 0,25X = 0,3(100+X)
0,05X = 10
X = 200

---> X/Y = 100/200 ; X+Y = 300 ---> X/(X+G) = 100/300 = 33,3% (B)

I've solved it this way, I hope it will be useful for some of you......
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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 %
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]  04 Jun 2014, 23:44
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Using Alligation method
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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?
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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%
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Re: Seed mixture X is ryegrass and bluegrass [#permalink]  10 May 2012, 07:10
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Final Requirement is to calculate "X/(X+Y) * 100"

We have the information about the quantity by weight of ryegrass in X,Y and also the quantity by weight of ryegrass after mixing i.e in X+Y.

So the equation satisfying the above mentioned criteria would be :

(40/100)*X+(25/100)*Y=(30/100)*(X+Y)

Solving this will lead us to the result...

2X=Y

So, the weight percent of X :
(X/X+Y)*100 = (X/X+2X)*100 = 33.33%
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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 ?

(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %

[Reveal] Spoiler:
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

Last edited by Bunuel on 26 Oct 2013, 02:23, edited 3 times in total.
Edited the question and added the OA. TOPIC LOCKED
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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 %

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.
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Re: Mixture Problem [#permalink]  17 Jul 2012, 20:16
It might be easier to think of things in terms of their weight. Assume that you're holding 100kg of the mixture. Of this 100kg, you have a certain amount of kilograms of X, and a certain amount of kilograms of Y. That's where you begin with the X + Y = 1. Note that you also know that there is 30kg of ryegrass in your mixture.

Each kilogram of X would consist of 400g of ryegrass and 600g of bluegrass, while each kilogram of Y would comprise 250g of ryegrass and 750g of fescue. If you want to know how many kilograms of X the mixture contains, you need to work out how you can arrive at 30kg of ryegrass, with exactly 70kg of other stuff (bluegrass + fescue) in the mix. That's what the rest of the working is referring to.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]  08 Mar 2013, 23:05
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RG detail table:

X----------------------------------Y
40%------------------------------25%
-----------------30%--------------
30-25 = 5------------------40-30 = 10

RG ratio 5:10 = 1:2

Therefore, % of X in new mixture = 1/3 = 33.3%
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Re: Seed mixture X is 40 percent ryegrass and 60 percent   [#permalink] 08 Mar 2013, 23:05

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