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Seed mixture X is 40 percent ryegrass and 60 percent bluegra

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Seed mixture X is 40 percent ryegrass and 60 percent bluegra [#permalink]  25 Jul 2010, 12:49
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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 %

OPEN DISCUSSION OF THIS QUESTION IS HERE: seed-mixture-x-is-40-percent-ryegrass-and-60-percent-60580.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 26 Oct 2013, 02:29, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Mixture problem using table method [#permalink]  25 Jul 2010, 13:28
Why do you need tables to solve? Don't use a more difficult method when a very easy one is right in front of you.

You know it's not D, because that would actually give you 32.5%, not 30%. It might be 40%, but it's not as easy to test as 33.33% is.

If you figure you have 100 lbs of X, then for X to be 33.33% of the end mixture, there needs to be 200 lbs of Y. So for 100 lbs of X you get 40 lbs of ryegrass, and then 25% of 200 lbs of Y = 50, so a total of 90 lbs of ryegrass and 300 lbs of mixture. 90/300 = 30%. We have a winner.

rxs0005 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 %

Can some one solve this mixture problem using Tables

thanks

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Re: Mixture problem using table method [#permalink]  25 Jul 2010, 13:58
---------------->Ryegrass
X--------------> 40%
Y--------------> 25%
M(mixture)---->30%

so
0.4X + (M-X)0.25 = 0.3M
0.15X = 0.05M
X = 1/3M

X = 33.33% of M
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Re: Mixture problem using table method [#permalink]  24 Aug 2010, 15:37
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Try solving this using Alligation (weighted mean).

Qnty.(cheaper)/Qnty.(dearer)=(dearer-mean)/(mean-cheaper)

Thus Qy/Qx=(40-30)/(30-25)=2

Therefore Qy=2 Qx and Qy+Qx = 100% (total). Hence Qx = 33.33%

I have encountered loads of problems of this kind so try to learn this approach if it is an option.
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Re: Mixture problem using table method [#permalink]  15 Oct 2010, 07:58
1
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simpl..
.4x+.25y=.3x+.3y
x/y = .5...
x/(x+y) = .5/1.5
hence 33.33%
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Re: Mixture problem using table method [#permalink]  17 Oct 2010, 19:12
zuperman wrote:
Try solving this using Alligation (weighted mean).

Qnty.(cheaper)/Qnty.(dearer)=(dearer-mean)/(mean-cheaper)

Thus Qy/Qx=(40-30)/(30-25)=2

Therefore Qy=2 Qx and Qy+Qx = 100% (total). Hence Qx = 33.33%

I have encountered loads of problems of this kind so try to learn this approach if it is an option.

Totally agree. It reminded me of my CAT days . This was one quick fix formula.
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Re: Mixture problem using table method [#permalink]  26 Oct 2013, 01:01
Hello from the GMAT Club BumpBot!

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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegra [#permalink]  26 Oct 2013, 02:30
Expert's post
OPEN DISCUSSION OF THIS QUESTION IS HERE: seed-mixture-x-is-40-percent-ryegrass-and-60-percent-60580.html
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegra   [#permalink] 26 Oct 2013, 02:30
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