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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]

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20 Apr 2015, 12:36
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.

Another such question can be found here -

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

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09 Nov 2015, 02:03
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%)

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.

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

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22 Dec 2015, 18:13
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, 19:54
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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29 Apr 2016, 09:38
1
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%

This is a great problem because we can solve it using equations and substitution, or we can employ some estimation because the equation we create is a weighted average. Let’s first do the equation-substitution method.

Equation-Substitution Method

We are given information about ryegrass, bluegrass, and fescue. However, we are told the mixture of X and Y is 30 percent ryegrass. Thus, we only care about the ryegrass, so we can ignore fescue or bluegrass.

We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the weight of the combined mixture is 30% ryegrass. With x representing the total weight of mixture X, and y representing the total weight of mixture Y, we can create the following equation:

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

We can multiply the entire equation by 100, and we have:

40x + 25y = 30x + 30y

10x = 5y

2x = y

The question asks what percent of the weight of the mixture is x.

We can create an expression for this:

x/(x+y) * 100 = ?

Since we know that 2x = y, we can substitute in 2x for y in our expression. So we have:

x/(x+2x) * 100 = x/(3x) * 100 = 1/3 * 100 = 33.33%

Estimation Method

This is a method I would suggest using only if you are short on time.

We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the overall weight of the mixture is 30% ryegrass.

Since 30% is closer to 25% than it is to 40%, we know that there is more of mixture Y than there is of mixture X, but not by a large amount. So let’s analyze our answer choices.

A) 10%

This tells us that the mixture is made up of 10% x and 90% y. That discrepancy is much too large to be correct, based on the overall weight of the mixture.

B) 33 1/3%

This tells us that the mixture is made up of 33 1/3% x and 66 2/3% y. This could be the answer.

C) 40%

This tells us that the mixture is made up of 40% x and 60% y. This could be the answer.

D) 50%

This tells us that the mixture is made up of 50% x and 50% y. Since we know that y is weighted more heavily in the mixture, we know this cannot be the answer.

E) 66.67%

This tells us that the mixture is made up of 66 2/3% x and 33 1/3% y. Since we know that y is weighted more heavily in the mixture, we know this cannot be the answer.

So now we have our answer choices narrowed down to B and C. However, there is something interesting about answer choice B. Notice that if we were to add together answer choice B (33 1/3%) and answer choice E (66 2/3%), our sum would be 100%. These are strategic answer choices, because the GMAT is hoping that if we make a mistake in our calculations we will determine that x represents 66 2/3% of the mixture rather than 33 1/3% of the mixture. Since answer choice C (40%) does not have a corresponding trap answer there is a higher likelihood that the answer is B (33 1/3%) than it is C (40%).

As mentioned at the beginning, the surest solution technique is to create an equation to solve it and actually complete the math. However, if you are short on time or have trouble setting up the equation, estimation is a method you can employ.

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

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02 Jul 2016, 07:17
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:

Balance Method:

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

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05 Jan 2017, 15:05
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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Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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31 Jan 2017, 17:10
Another way with algebra is to think about it more conceptually. An easier way to look at it is as a weighted average of both mixtures. The weighted avg. is closer to the average of the larger group. Since X is 40% rye and Y is 25% rye and the new amount (30%) is closer to Y than to X, there must be more X in the new mixture. X is twice as far from the value as Y is so the proportion of Y:X is going to be 2:1. Hence the answer 33.33%.

If the new mixture was 28% rye and Y is 4 times the distance from 28 as X. The proportion of Y:X would be 4:1 notice how as the distance moves closer to one value relative to the other the proportions increase dramatically!

Another way without algebra is to think about the mixture more conceptually. X is 40% rye and Y is 25% rye. Since the overall mixture of 30% is closer to 25% than to 40% rye, there must be more Y than X in the overall mixture. The weighted average is actually a concept tested here and since X is twice as far from 30% (40% - 30% = 10%) as Y (25% - 30% = 5%) is, the proportions will line up as Y:X ---> 2:1.

If the new mixture had been 28% the new proportion of Y:X would be 4:1 since 12 is four times the distance of 3 from the new value.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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20 May 2017, 06:33
Given:
Mixture X, 40 % ryegrass, 60% bluegrass by weight
Mixture Y 25% ryegrass, 75% fescue.

We need to find what is the % of X ryegrass in new mixture, where, total % of ryegrass is 30. At this point, we can focus of ryegrass only.

Based on the numbers given, we can form following equation:

0.40x + 0.25y = 0.3(x+y)

0.40x + 0.25y = 0.3x + 0.3y

0.1x = 0.05y

We can multiply everything by 100:

10x=5y
2x=y.
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Now let's form the equation for the question being asked: what is the percent of the weight of mixture X (part/whole):

x/(x+y)*100 = ?

x/(x+2x)*100 = x/3x*100 = 1/3 = 33.33%.=> answ. B.
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Re: Seed mixture X is 40% ryegrass and 60% bluegrass by weight; seed mixtu  [#permalink]

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16 Oct 2017, 16:57
carollu wrote:
Seed mixture X is 40% ryegrass and 60% bluegrass by weight; seed mixture Y is 25% 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)33.33%
c)40%
d)50%
e)66.67%

We are given information about ryegrass, bluegrass, and fescue. However, we are told the mixture of X and Y is 30 percent ryegrass. Thus, we only care about the ryegrass, so we can ignore fescue and bluegrass.

We are given that seed mixture X is 40% ryegrass and that seed mixture Y is 25% ryegrass. We are also given that the weight of the combined mixture is 30% ryegrass. With x representing the total weight of mixture X, and y representing the total weight of mixture Y, we can create the following equation:

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

We can multiply the entire equation by 100:

40x + 25y = 30x + 30y

10x = 5y

2x = y

The question asks what percentage of the weight of the mixture is x.

We can create an expression for this:

x/(x+y) * 100 = ?

Since we know that 2x = y, we can substitute in 2x for y in our expression. So, we have:

x/(x+2x) * 100 = x/(3x) * 100 = 1/3 * 100 = 33.33%

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

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28 Oct 2017, 23:49
40 :::::::::: 25
::::::30
30-25: 40-30
5:10
ratio => 5/15 => 33.33%
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Re: Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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31 Oct 2017, 02:10
I assumed values

x = 100, so reygrass = 40 and bluegrass =60
y = 200, so ryegrass = 50 and fescue = 150

what percent of the weight of the mixture is X?

40/300 * 100 = 40/3 - its wrong. where am i going wrong here?
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Re: Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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31 Oct 2017, 04:46
santro789 wrote:
I assumed values

x = 100, so reygrass = 40 and bluegrass =60
y = 200, so ryegrass = 50 and fescue = 150

what percent of the weight of the mixture is X?

40/300 * 100 = 40/3 - its wrong. where am i going wrong here?

you cant assume values..
why u have not assumed y =100 or y=300 different values -> different answers.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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05 Jan 2018, 13:31
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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%

This looks like a job for weighted averages!

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Mixture X is 40 percent ryegrass
Mixture Y is 25 percent ryegrass

Let x = the PERCENT of mixture X needed (in other words, x/100 = the proportion of mixture X needed)
So, 100-x = the PERCENT of mixture Y needed (in other words, (100-x)/100 = the proportion of mixture Y needed)

Weighted average of groups combined = 30%

Now take the above formula and plug in the values to get:
30 = (x/100)(40) + [(100-x)/100](25)
Multiply both sides by 100 to get: 30 = 40x + (100-x)(25)
Expand: 3000 = 40x + 2500 - 25x
Simplify: 3000 = 15x + 2500
So: 500 = 15x
Solve: x = 500/15
= 100/3
= 33 1/3

So, mixture X is 33 1/3 % of the COMBINED mix.

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

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05 Jan 2018, 16:11
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%

let x=% of X in mixture
y=% of Y in mixture
.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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07 Jan 2018, 03:43
VeritasPrepKarishma niks18

Quote:
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.

I went through your blog post here
wherein you mentioned to put lower no of LHS on scale line. Accordingly if W1=40% and W2=25% and combined average is 25%
I am getting W1/W2 =10/5 or 2/1. Can you add your two cents?
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Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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07 Jan 2018, 05:36
VeritasPrepKarishma niks18

Quote:
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.

I went through your blog post here
wherein you mentioned to put lower no of LHS on scale line. Accordingly if W1=40% and W2=25% and combined average is 25%
I am getting W1/W2 =10/5 or 2/1. Can you add your two cents?

There is a very simple way to solve this problem. so if the $$x$$ & $$y$$ are the weights of the mixtures, we need to find

$$\frac{x}{x+y}*100$$---------------------(1)Essentially you need a relationship between $$x$$ & $$y$$.

from the question stem it is clear that the only connecting factor here is ryegrass whose wight in $$x$$, $$y$$ and combined mixture is given. So we have a relationship here -

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

solve this to get $$y=2x$$. substitute this in our equation (1) to get

$$\frac{x}{x+2x}*100=\frac{1}{3}*100=33\frac{1}{3}$$%

------------------------------------------------------

Essentially this is same as the weighted average method, where we are given weights of ryegrass within x & y and it's weighted average in the mixture

(40%x+25%y)/(x+y)=30%
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Re: Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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08 Jan 2018, 04:02
VeritasPrepKarishma niks18

Quote:
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.

I went through your blog post here
wherein you mentioned to put lower no of LHS on scale line. Accordingly if W1=40% and W2=25% and combined average is 25%
I am getting W1/W2 =10/5 or 2/1. Can you add your two cents?

W1 and A1 need to represent the same quantity and W2 and A2 need to represent the same quantity. When using the weight average formula, you can take either to be ingredient 1.

Seed mixture X is 40 percent ryegrass by weight - Say this represents W1 and A1
seed mixture Y is 25 percent ryegrass - Say this represents W2 and A2

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (25 - 30)/(30 - 40) = 1:2

Whatever you do keep the name consistent. Check out this post to understand: https://www.veritasprep.com/blog/2017/1 ... -averages/
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Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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08 Jan 2018, 12:36
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%

Hello everyone :)

ok here is my "correct" soloution

i used formula of weighted average:

$$\frac{W1}{W2}$$ =$$\frac{Avg2 - Avg}{Avg-Avg1}$$

X mixture = .40ryegrass + .60bluegrass
Y mixture = .25ryegrass + .75fescue

combined ryegrass .40ryegrass + .25ryegrass = 30%

so :-) i asked myself a question which one X.40 or Y.25 has more amount of ryegrass ? so the one that is closer to combined Average .30, has more amount of rygrass - hence Y.25

So, W1 = Y.25 and W2 =X.40 --->$$\frac{W1}{W2}$$ = $$\frac{40 - 30}{30-25}$$ =$$\frac{10}{5}$$ = 2

So where am i wrong ? i did exactly, as Bunuel explained here https://gmatclub.com/forum/weighted-ave ... 06999.html

perhaps you friends can help chetan2u , niks18
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Re: Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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09 Jan 2018, 05:56
dave13 wrote:
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%

Hello everyone

ok here is my "correct" soloution

i used formula of weighted average:

$$\frac{W1}{W2}$$ =$$\frac{Avg2 - Avg}{Avg-Avg1}$$

X mixture = .40ryegrass + .60bluegrass
Y mixture = .25ryegrass + .75fescue

combined ryegrass .40ryegrass + .25ryegrass = 30%

so i asked myself a question which one X.40 or Y.25 has more amount of ryegrass ? so the one that is closer to combined Average .30, has more amount of rygrass - hence Y.25

So, W1 = Y.25 and W2 =X.40 --->$$\frac{W1}{W2}$$ = $$\frac{40 - 30}{30-25}$$ =$$\frac{10}{5}$$ = 2

So where am i wrong ? i did exactly, as Bunuel explained here https://gmatclub.com/forum/weighted-ave ... 06999.html

perhaps you friends can help chetan2u , niks18

Hi..

you have gone wrong in taking base of weighted average..
you are looking for % of entire mixture and that will be 40-25..
x will be less as 30 is closer to y(25) so x will be 30-25..
ans $$\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33$$%

Hope it helps
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