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Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig

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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig  [#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.

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

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29 Apr 2016, 09:38
4
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%

The answer is B.

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.

The answer is B.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig  [#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 bluegrass by weig  [#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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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig  [#permalink]

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16 Oct 2017, 16:57
1
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 bluegrass by weig  [#permalink]

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05 Jan 2018, 13:31
Top Contributor
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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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig  [#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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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig  [#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 bluegrass by weig  [#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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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig  [#permalink]

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28 Jul 2018, 07:48
chintanpurohit wrote:
E.66 2/3%

solve for 25x + 40 [1-x] = 30

so, x = 2/3 = 66 2/3%

How did u get 1-x?9
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Re: Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weig  [#permalink]

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04 Aug 2018, 20:24
S1937 wrote:
chintanpurohit wrote:
E.66 2/3%

solve for 25x + 40 [1-x] = 30

so, x = 2/3 = 66 2/3%

How did u get 1-x?9

Hi S1937,

It turns out that [1-x] is an error, if x is meant to represent the fraction of the final mixture made up by X. Instead, [1-x] should be the fraction of the final mixture that is made up by Y, and the equation should read:

25[1-x] + 40x = 30

Then you get x = 1/3 = 33 1/3%, and arrive at the correct answer, B.

I also wanted to let everyone know that I'm starting a new project to go through all of the Quant questions in the 2019 Official Guide and share my suggested "GMAT Timing Tips" for each one: that is, what could you notice or do to be able to answer each question more efficiently? For this question, my GMAT Timing Tips are:

Weighted average mapping strategy (also known as the tug of war, and similar to the "Balance Method" also described in this thread): Use the weighted average mapping strategy to find the ratio of X to Y. This is a shortcut for the algebra, and can be a great way to save valuable time on the GMAT.

I'm also planning to create video solutions demonstrating how to use my GMAT Timing Tips for each of these questions. This will take a while, but I'll start with the those that I get requests for first. If you would me to create a video solution for this question, please go to the page that I have created for this question and vote for or share it: Seed mixture X is 40 percent ryegrass and 60 percent...

You can also ask me to answer a question about this GMAT practice question by asking it in this thread, or by using the question form on the page above. Please let me know if you have any questions!
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Re: Seed mixture X is 40 percent ryegrass and 60 percent  [#permalink]

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Re: Seed mixture X is 40 percent ryegrass and 60 percent   [#permalink] 31 Jul 2019, 02:22

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