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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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13 Feb 2004, 11:02
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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 ?
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shubhangi

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13 Feb 2004, 11:28
x = 40r + 60b
y = 25r + 75f
x+y = 65r+60b+75f

now 65 gms is 30% by weight then total weight of the mixure is 65/0.3 gms
out of which 100 grms is contributed by mixture x
so ratio of x/x+y = 100*0.3 / 65 = 30/65 approx 46%
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13 Feb 2004, 11:31

Do you have any problems in showing your work
Woudl you like if someone just gives you the answer for your questions?
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13 Feb 2004, 11:35
anandnk wrote:

Do you have any problems in showing your work
Woudl you like if someone just gives you the answer for your questions?

I just mistook 60 for 65. Then after seeing your answer. I deleted mine. So, the answer should be (30/65)*100 which is approximately 46.
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13 Feb 2004, 11:38
anandnk wrote:

Do you have any problems in showing your work
Woudl you like if someone just gives you the answer for your questions?

I do have problems in giving the workout because of lack of confidance. Till i know mine is the right answer i try not to confuse them.
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13 Feb 2004, 11:50
it is from one of the oldest probl in this forum..
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shubhangi

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13 Feb 2004, 12:35
hi,

I believe kpadma needs to be kicked ( Just kidding ) for not showing his work.

Here is my anlysis
Let us say they are both mixed in equal proprtions
X contributes 100 gms
Y contributes 100 gms
X+Y = 200 gms
Out X we get 40 gms of R and out of Y we get 25 gms of R
so total = 65 gms
now 65/(200) = 32.5% Since the stem says 30% we can confidently say that they are not mixed in equal proprtions

Let us say Y is added two times making it contribute
50 gms of R and 150 gms of F
then total R = 50+40 = 90
and X+Y = 100+200 = 300

Now net R / total weight = 90/300 = 30% which is what the stem says

So X/(X+Y) = 100(100+200) = 33.3%

Can some one reverse engineer this and show it by set of equations ?
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I did it this way.. [#permalink]

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13 Feb 2004, 13:15
Let z be the fraction (U can assume in % terms as well) of mixture X.

1-z is the fraction of mixture Y.

(40*z + 25(1-z)) / (z+1-z) = 30

15z = 5,

z = 1/3 => 33.33% of X
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13 Feb 2004, 13:17
good show mantha
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13 Feb 2004, 15:00
anandnk wrote:

Do you have any problems in showing your work
Woudl you like if someone just gives you the answer for your questions?

I usually post the method after few hours or a day, if my approach, IMO,
is better than other posts. This I learned it from our Guru, Akamai

тАЬThe harder one tries to solve a problem, the longer he retains"

There are two reasons.
1. It allows other to think in new direction.
2. In office, it many not be possible to type the steps

I'll try to post answers for the difficult problems, as soon as I get time
if no other post has better method.

This is a collaborative place, where each learns from others.
I owe a lot to you all, because I learned a lot from you !
My method for solving the question in the next post!

Last edited by kpadma on 13 Feb 2004, 15:18, edited 1 time in total.
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13 Feb 2004, 15:17
Good one
Ans: 33.3%

Steps.
-------

Mixture X: 0.4R + 0.6G
Mixture Y: 0.25R + 0.75G
Mxsture XY: 0.3R + 0.7G

G - All other mateial (let us say garbage)

Let us say, A pounds of X and B pounds of Y are mixed to
get (A+B) pounds of XY mixture.

0.4A + 0.25B = 0.3(A+B)
A/B = 1/2
A/(A+B) = 1/3 = 33.3%

It took more than 5 minutes to solve this
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13 Feb 2004, 17:03
kpadma's solution is easy to follow. post more mixture problems shubhangi.
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