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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]

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New post 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?

A. 10%
B. 33 1/3%
C. 40%
D. 50%
E. 66 2/3%

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Re: 223 Percentage  [#permalink]

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New post 23 Feb 2011, 19:21
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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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Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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New post 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:

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: 240. PS.Mixture  [#permalink]

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New post 27 Feb 2008, 01:41
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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]

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New post 27 Feb 2008, 19:06
1
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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Re: Seed mixture X is ryegrass and bluegrass  [#permalink]

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New post 28 Jan 2010, 04:47
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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]

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New post 28 Jan 2010, 22:57
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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]

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

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New post 05 Feb 2011, 05:27
2
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% ryegrass and 60% bluegrass by weight; seed mixtu  [#permalink]

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New post 09 May 2011, 06:23
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1
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 ryegrass and bluegrass  [#permalink]

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New post 10 May 2012, 08:10
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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]

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New post Updated on: 26 Oct 2013, 03:23
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=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

Originally posted by debabrata44 on 17 Jul 2012, 11:34.
Last edited by Bunuel on 26 Oct 2013, 03:23, edited 3 times in total.
Edited the question and added the OA. TOPIC LOCKED
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Re: Mixture Problem  [#permalink]

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New post 17 Jul 2012, 11:43
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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]

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New post 17 Jul 2012, 21:16
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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]

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New post 09 Mar 2013, 00:05
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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]

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New post 09 Mar 2013, 00:25
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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]

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New post 21 Jul 2013, 14:21
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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]

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

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New post 03 Jun 2014, 21:45
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 100

If 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  [#permalink]

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Re: Seed mixture X is 40 percent ryegrass and 60 percent &nbs [#permalink] 05 Jun 2014, 00:44

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