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PS-Mixture

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Senior Manager
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PS-Mixture [#permalink]

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New post 25 Nov 2008, 21:49
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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 1/3%
(C) 40%
(D) 50%
(E) 66 2/3%

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Re: PS-Mixture [#permalink]

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New post 25 Nov 2008, 22:07
I believe I have seen this before. was tough

what we need here is to put together the correct eq

combined mixture percentage = sum of percentages in both

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

0.1x = 0.05y

means 10x = 5y and y = 2x

Q is asking for (x/x+y) and it is 1/3 which is B

The Q threw me out of loop with unnecessary equations such as x = 0.4 R + 0.6 B and y = .25R + .75F

All of a sudden we have 3 variables and 2 equations. Dang! I went crazy!

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Manager
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Re: PS-Mixture [#permalink]

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New post 26 Nov 2008, 04:04
Hi,
let us assume X and Y were mixed in the ration a:b.
we can write a equation:-

\(2/5*a/(a+b) + 1/4*b/(a+b)\)
------------------------------------------- \(=3/7\)
\(3/5*a/(a+b) + 3/4*b/(a+b)\)

Solving we get, \(a:b=1/2\)
hence \(a:(a+b)=1:3\).

(A)

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Re: PS-Mixture [#permalink]

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New post 26 Nov 2008, 07:43
icandy wrote:

The Q threw me out of loop with unnecessary equations such as x = 0.4 R + 0.6 B and y = .25R + .75F

All of a sudden we have 3 variables and 2 equations. Dang! I went crazy!


Same here, icandy!

Agree with B.. same approach.

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Re: PS-Mixture [#permalink]

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New post 26 Nov 2008, 07:59
The way I approached this is:
assume out of the 30% ryegrass, x% is from X
hence, amount of bluegrass in the mixture = 3x/2
similarly, (30-x)% is from Y
hence, amount of fescue = 3(30-x)

x + 3x/2 + (30-x) + 3(30-x) =100
solve for x => 40/3
3x/2 => 20

20% of the mixture is bluegrass, therefore, total for X=40/3+20=100/3 = 33 1/3
B

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Re: PS-Mixture [#permalink]

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New post 26 Nov 2008, 14:56
twilight,

can you explain your approach clearly?

Thanks

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Re: PS-Mixture [#permalink]

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New post 26 Nov 2008, 15:51
Sure,
R=Ryegrass, B=Bluegrass, F=Fescue
X: 40% R, 60% B
Y: 25% R, 75% F
Mixture(lets call it M): 30% R, some B, some F
Question: What% of M is X

Since M is mixture, and ryegrass is M is from both X and Y,
Hence, the 30% of Ryegrass, will be the sum of ryegrass which comes from X and from Y.
Assume that the % of ryegrass contributed by X is x.
Total ryegrass = 30%
% from X= x%
Therefore, percentage from Y= (30-x)%

Now, in X the ratio of R:B = 40:60
Hence, if R=x, B=(60/40)x=3x/2

Similarly, in Y the ratio of R:F = 25:75
Hence, if R=30-x, F=(75/25)(30-x) = 3(30-x)

Adding all gives you 100%, so,
x + 3x/2 + (30-x) + 3(30-x) =100
solve for x => 40/3
B= 3x/2 => 20

To find total % of X in M, you have to add amount of B and amount of R from X
= 20 + 40/3 = 33.33

Hence B

Hope that makes sense!

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Re: PS-Mixture   [#permalink] 26 Nov 2008, 15:51
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