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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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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. \(33\frac{1}{3}\)% C. 40% D. 50% E. \(66\frac{2}{3}\)%
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Originally posted by sondenso on 26 Feb 2008, 18:46.
Last edited by Bunuel on 18 Jun 2019, 20:44, edited 1 time in total.
Renamed the topic and edited the question.




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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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23 Feb 2011, 18:21
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/toughds105651.html#p828579) Hence X is 33.33% in the mixture.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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17 Aug 2009, 23:00
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 (100X)g.
So we can now equate the parts of the ryegrass in the mixture as:
0.4X + 0.25(100X) = 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)
0.4x0.3x = 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: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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27 Feb 2008, 00:41
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 (1x) = 30 solve for x, x = 33.33%



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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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27 Feb 2008, 18:06
the answer is E.66 2/3%
solve for 25x + 40 [1x] = 30
so, x = 2/3 = 66 2/3%



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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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28 Jan 2010, 03:47
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(Mx) = 0.3M .4x + 0.25M  0.25x = 0.3M .15x = .05M x/M = 1/3
Hence ans = 33.33%.



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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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28 Jan 2010, 21:57
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)
0.4x0.3x = 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%



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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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24 Sep 2010, 22:41
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%



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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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05 Feb 2011, 04:27
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



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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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09 May 2011, 05:23
x  wt of Mix X added y  wt of Mix Y added 0.4x + 0.25y = 0.3(x+y) 0.1x = 0.05y x/y = 5/10 = 1/2 y/x = 2 (y+x)/x = 3 => x/(y+x) * 100 = 100/3 = 33.33%
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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10 May 2012, 07:10
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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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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Updated on: 26 Oct 2013, 02:23
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=1X
(4/10)X+(1/4)Y = (3/10)(X+Y)
(4/10)X+(1/4)(1X) = (3/10)(x+(1X))
get common denominator
(16x+(1010x))/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
Originally posted by debabrata44 on 17 Jul 2012, 10:34.
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: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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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=1X
(4/10)X+(1/4)Y = (3/10)(X+Y)
(4/10)X+(1/4)(1X) = (3/10)(x+(1X))
get common denominator
(16x+(1010x))/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: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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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 bluegrass by weig
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08 Mar 2013, 23:05
RG detail table: XY 40%25% 30% 3025 = 54030 = 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 bluegrass by weig
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08 Mar 2013, 23:25
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/
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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21 Jul 2013, 13:21
........,,,,,,...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 bluegrass by weig
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21 Jul 2013, 17:43
1. x and y should have been mixed in the ratio, 3025 / 4030 = 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 bluegrass by weig
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03 Jun 2014, 20:45
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% Quote: In this my approach was that let weight of X be 100 and weight of Y be 100
Now For X, ryegrass is 40% of 100=40
and for total mixture it will be 30%of 200=60
To calculate the percentage weight from mixture X why it is wrong 40/60 *100
Your first assumption is incorrect here: In this my approach was that let weight of X be 100 and weight of Y be 100If you are assuming that the weights are 100 each and the combined weight is 200, what is left to find? X would be 1/2 of the mixture 100/200. What you have to find is the ratio of the weights in which they were mixed. The ratio may not be 1:1 as you have assumed here. X has 40% ryegrass and Y has 25% ryegrass. Total mix has 30% ryegrass. So X/Y = (25  30)/(30  40) = 1/2 So X by weight is 1 part (say 100 gms) and Y is 2 parts (say 200 gms). So X is 1/3 of the total mix i.e. 33.33%
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig
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04 Jun 2014, 23:44
Using Alligation method
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