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Seed mixture X is 40 percent ryegrass and 60 percent

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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 11 Nov 2013, 20:01
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%



Its very simple ..

30/100(x+y)= 40/100(x) + 25/100(y)
after solving we'll get, y=2x ... and we know x + y = 100(Total mixture)

x = 33.33
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 18 Apr 2014, 20:49
Another Approach:

X : 0.40 R and 0.60 B

Y: 0.25 R and 0.75 F

Mixture:

0.40 X + 0.25 Y = 0.30 (X+Y)

Y=2X

We need to know what is X/(X+Y)

1/3
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 03 Jun 2014, 20:45
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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: 223 Percentage [#permalink] New post 03 Jun 2014, 21:47
VeritasPrepKarishma wrote:
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
Hence X is 33.33% in the mixture.



Hi Karishma,

So cann't we use plugging in for these types of questions like assuing the weight be anything etc. Like what i did was 100 and 100 for x and y which was wrong as we cannot assume both of the weights to be equal . Hence for the questions which require calculation for wieighted percentage, we will have to go by equations.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 04 Jun 2014, 23:44
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 28 Aug 2014, 08:23
This is similar to question 185 for those seeking a similar problem. That said, this question (about seed mixtures) is a bit more challenging.
Re: Seed mixture X is 40 percent ryegrass and 60 percent   [#permalink] 28 Aug 2014, 08:23
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