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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 13 Feb 2004, 10:02
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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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 [#permalink] New post 13 Feb 2004, 10:28
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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 [#permalink] New post 13 Feb 2004, 10:31
hi kpadma and geethu,

Do you have any problems in showing your work :lol:
Woudl you like if someone just gives you the answer for your questions?
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 [#permalink] New post 13 Feb 2004, 10:35
anandnk wrote:
hi kpadma and geethu,

Do you have any problems in showing your work :lol:
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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 [#permalink] New post 13 Feb 2004, 10:38
anandnk wrote:
hi kpadma and geethu,

Do you have any problems in showing your work :lol:
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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 [#permalink] New post 13 Feb 2004, 10:50
it is from one of the oldest probl in this forum..
the answer matches with Kpadma.. but anand's method seems right..though :?
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 [#permalink] New post 13 Feb 2004, 11:35
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% :lol:


Can some one reverse engineer this and show it by set of equations ? :?:
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I did it this way.. [#permalink] New post 13 Feb 2004, 12:15
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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 [#permalink] New post 13 Feb 2004, 12:17
good show mantha :lol:
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 [#permalink] New post 13 Feb 2004, 14:00
anandnk wrote:
hi kpadma and geethu,

Do you have any problems in showing your work :lol:
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

His advice (not in verbatim)
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 :roll:

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, 14:18, edited 1 time in total.
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 [#permalink] New post 13 Feb 2004, 14:17
kpadma wrote:
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 :cry:
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 [#permalink] New post 13 Feb 2004, 16:03
kpadma's solution is easy to follow. post more mixture problems shubhangi.
  [#permalink] 13 Feb 2004, 16:03
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