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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] New post 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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Re: 240. PS.Mixture [#permalink] New post 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%
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Re: 240. PS.Mixture [#permalink] New post 27 Feb 2008, 18:06
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the answer is
E.66 2/3%

solve for 25x + 40 [1-x] = 30

so, x = 2/3 = 66 2/3%
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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 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: Seed mixture X is ryegrass and bluegrass [#permalink] New post 28 Jan 2010, 03:47
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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%.
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Re: Seed mixture X is ryegrass and bluegrass [#permalink] New post 28 Jan 2010, 21:57
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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%
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Re: Seed mixture X is ryegrass and bluegrass [#permalink] New post 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%
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Re: Seed mixture X is ryegrass and bluegrass [#permalink] New post 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] New post 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] New post 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: 223 Percentage [#permalink] New post 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.
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Re: Seed mixture X is ryegrass and bluegrass [#permalink] New post 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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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 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 %

Answer explanation given as :
[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] New post 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.
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Re: Mixture Problem [#permalink] New post 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] New post 08 Mar 2013, 23:05
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] New post 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.

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 [#permalink] New post 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 [#permalink] New post 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] New post 26 Oct 2013, 06:05
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%


The answer is B

Here is how I would solve this question. The focus area is Ryegrass

Background of this method.

if A1 is the avg of group1 containing n1 elements and
if A2 is the avg of group2 containing n2 elements
and let AW be the avg of the group
then AW=(n1A1)+(n2A2)/n1+n2

Another key thing to note here is if A1<A2
then A1<Aw<A2
=A2-AW:AW-A1
=n1:n2

Hence 40 : 25
=A2:A1
=A2-Aw:Aw-A1
=40-30:30-25
=2:1
=33.33% (100/3)
Re: Seed mixture X is 40 percent ryegrass and 60 percent   [#permalink] 26 Oct 2013, 06:05
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