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# Seed mixture X is 40 percent ryegrass and 60 percent

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17 Jul 2012, 10:43
debabrata44 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 %

1=X+Y --> Y=1-X

(4/10)X+(1/4)Y = (3/10)(X+Y)

(4/10)X+(1/4)(1-X) = (3/10)(x+(1-X))

get common denominator

(16x+(10-10x))/40 = (3/10)

6x+10=12

6x=2

x=2/6 = 1/3 => 33.3%

Can anyone pls expain me how they are taking 1=X+Y.

Pls response. Thanks a lot

The question asks "what percent of the weight of this mixture is X".
If we denote by X the percentage of the mixture which is of type X and by Y the percentage of the mixture which is of type Y, then together they must give 100% = 1.
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17 Jul 2012, 20:16
It might be easier to think of things in terms of their weight. Assume that you're holding 100kg of the mixture. Of this 100kg, you have a certain amount of kilograms of X, and a certain amount of kilograms of Y. That's where you begin with the X + Y = 1. Note that you also know that there is 30kg of ryegrass in your mixture.

Each kilogram of X would consist of 400g of ryegrass and 600g of bluegrass, while each kilogram of Y would comprise 250g of ryegrass and 750g of fescue. If you want to know how many kilograms of X the mixture contains, you need to work out how you can arrive at 30kg of ryegrass, with exactly 70kg of other stuff (bluegrass + fescue) in the mix. That's what the rest of the working is referring to.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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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%

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]

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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]

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03 Jun 2014, 20:45
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 25 Feb 2014 Posts: 12 Followers: 0 Kudos [?]: 3 [0], given: 2 Re: 223 Percentage [#permalink] ### Show Tags 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. Current Student Joined: 08 Feb 2014 Posts: 206 Location: United States Concentration: Finance GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking) Followers: 2 Kudos [?]: 77 [0], given: 145 Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] ### Show Tags 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. Intern Joined: 06 Oct 2013 Posts: 48 Followers: 1 Kudos [?]: 29 [0], given: 33 Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] ### Show Tags 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 Intern Joined: 12 Jun 2014 Posts: 4 Followers: 0 Kudos [?]: 0 [0], given: 162 Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] ### Show Tags 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. Manager Joined: 06 Jul 2011 Posts: 132 Followers: 0 Kudos [?]: 71 [0], given: 240 Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] ### Show Tags 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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13678 [0], given: 222 Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink] ### Show Tags 09 Nov 2015, 01:03 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 http://gmatclub.com/forum/tough-ds-105651.html#p828579) Hence X is 33.33% in the mixture. Quote: Isn't the ratio for X:Y = 2:1 ? I do not understand simply how you are getting it as 1:2 ???? Because i was following all your posts for mixture problems, and the answer 1:2 seems to be contradicting the weighted average formula. How i am solving this is w1/ w2=(A2–Aavg)(Aavg–A1). Here i am taking Aavg as 30, A2= 25 A1 = 40 So 25-30/30-40 = 1/2 which is ratio for Y/X. so X/Y is 2/1. Its really really confusing for me which ratio to assign the equation to. I am getting many mixture problems wrong because of this. Can you please help clarify? Highlighted part is wrong. Use the formula as it is. X has 40% ryegrass and Y has 25% ryegrass to give 30% ryegrass in mixture w1/ w2=(A2–Aavg)(Aavg–A1) wX/wY = (AY - Aavg)/(Aavg - AX) wX/wY = (25 - 30)/(30 - 40) = 1/2 So wX/wY = 1/2. You get X:Y as 1:2 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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22 Dec 2015, 17:13
Simple and quick method as Karishma explained in this and other various posts:

$$\frac{Wx}{Wy} = \frac{30-25}{40-30} = \frac{5}{10} = \frac{1}{2}$$

Ratio of new mixture is 1:2, end result total is 3.

Solution X will be $$\frac{1}{3}$$ and Y $$\frac{2}{3}$$.

$$100 * \frac{1}{3} = 33\frac{1}{3}%$$

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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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28 Apr 2016, 18:54
Very simple and straight forward question, here is my scale method

Mix X : 40% Rye 60% BG
Mix Y : 25% Rye 75% Fescue

40%...........30%...........25%
.........10..............5........
.......further.....closer.....
.....x..................2x (double the weight of Mix X)
3x = Whole
1 x of whole = 1/3 of the weight
33 1/3 %
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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02 Jul 2016, 06:17
Here are a couple of video explanations for this question. The first goes over an algebraic approach. The second goes over a 'balance method' approach.

Algebraic Method: https://youtu.be/obHR1MewMhA
Balance Method: https://youtu.be/EkkOwiyrhhQ
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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02 Jul 2016, 09:43
.4x+.25y=.3(x+y)
x/y=1/2
x/(x+y)=1/3=33 1/3%
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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25 Sep 2016, 05:35
40/100 (x) + 25/100 (T-x) = 30/100 (T)

40/100 (x) - 25/100 (x) = 5/100 (T)

15/100 (x) = 5/100 (T) --> x = 0.33 (T)
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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03 Oct 2016, 21:06
x Y
40 25
30
(30-25) : (40-30)
5 : 10
1 : 2

so x is 1/(1+2) ie. 1/3 nothing but 33 1/3 percent
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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05 Jan 2017, 14:05
First of all, we don't care at about bluegrass or whatever fescue is. If we combine the mixtures to arrive at 30 percent ryegrass, we can just keep track of how we mix them according simply to the ryegrass; whatever else is in the mixture will follow accordingly and work out as it will. We just need to see how a mixture with 40 percent ryegrass and a mixture with 25 percent ryegrass combine to land at 30 percent.

What if you poured these 2 mixtures in equal quantities? Where would the ryegrass percentage land? It would of course land right in the middle. Equal parts of a 40 percent solution and a 25 percent solution will land us right in the middle, at 32.5 percent. However, in this problem that isn't where we land; we land at 30 percent. Think to yourself: if we mixed them evenly, the result would be a 32.5 percent solution and here we have 30, so which must there be more of? Which of the two mixtures, 25 or 40 percent, is the result closer to?

The 25 percent solution! 30 is closer to 25 than it is to 40, so there is more of the 25 percent solution, mixture Y. Even if you had no idea where to go from here, you can eliminate answers d and e. There has to be less of mixture X.

Now, let's return to thinking about that average mixture (because when you mix liquids of different salience, you're arriving at an average salience). Because 30 is twice as close to 25 as it is to 40 (5 away versus 10 away), we can reliably infer that there will be twice as much of the 25 percent mixture. That means it will be 2/3 Mixture Y (the 25 percent mixture) and 1/3 Mixture X (40 percent). Our answer is B.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent   [#permalink] 05 Jan 2017, 14:05

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# Seed mixture X is 40 percent ryegrass and 60 percent

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