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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 08 Mar 2013, 23:05
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RG detail table:

X----------------------------------Y
40%------------------------------25%
-----------------30%--------------
30-25 = 5------------------40-30 = 10

RG ratio 5:10 = 1:2

Therefore, % of X in new mixture = 1/3 = 33.3%
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 26 Oct 2013, 06:05
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%


The answer is B

Here is how I would solve this question. The focus area is Ryegrass

Background of this method.

if A1 is the avg of group1 containing n1 elements and
if A2 is the avg of group2 containing n2 elements
and let AW be the avg of the group
then AW=(n1A1)+(n2A2)/n1+n2

Another key thing to note here is if A1<A2
then A1<Aw<A2
=A2-AW:AW-A1
=n1:n2

Hence 40 : 25
=A2:A1
=A2-Aw:Aw-A1
=40-30:30-25
=2:1
=33.33% (100/3)
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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
Expert's post
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 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.
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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 28 Nov 2014, 09:17
Let resulting mixture is M.

30M=40x+25Y
70M=60X+75Y
Solving above equations for X in terms of M is , 4M=12X
so X=M/3 which is 33.33%

Ans.B
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 25 Jan 2015, 03:08
tejal777 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 %

METHOD 1:
Assuming the weight of the mixture to be 100g**, then the weight of ryegrass in the mixture would be 30g.
Also, assume the weight mixture X used in the mixture is Xg, then the weight of mixture Y used in the mixture would be (100-X)g.

So we can now equate the parts of the ryegrass in the mixture as:

0.4X + 0.25(100-X) = 30
0.4X + 25 - 0.25X = 30
0.15X = 5
X = 5/0.15 = 500/15 = 100/3

So the weight of mixture X as a percentage of the weight of the mixture
= (weight of X/weight of mixture) * 100%
= (100/3)/100 * 100%
= 33%

METHOD 2
wt. of 1st mixture = x
therefore concentration of ryegrass in 1st mix = 0.4x

wt. of 2nd mixture = y
therefore concentration of ryegrass in 2nd mix = 0.25y

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

0.4x-0.3x = 0.3y - 0.25y

0.1x=0.05y
or
2x=y

so if weight of x = 50grams
weight of y = 100 grams

total weight of mix = 150 grams

percentage of x in 150 grams of mix is 150*x/100 = 50

x = 50*100/150
x = 100/3
x = 33.3%



Thanx tejal777. Personally I found Method 2 very simple to understand.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] New post 20 Apr 2015, 11:36
Use this amazing formula to answer faster -

w1/w2 = (A2 – Aavg)/(Aavg – A1)

A1 = 40%
A2 = 25%
Aavg = 30%

Plugging in numbers, we get -

w1/w2 = 1/2
so So mixture X is 1/3 i.e. 33.33% of the mix.

Answer (B)

Another such question can be found here -

a-feed-store-sells-two-varieties-of-birdseed-brand-a-which-88125.html
Re: Seed mixture X is 40 percent ryegrass and 60 percent   [#permalink] 20 Apr 2015, 11:36

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