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555-605 Level|   Mixture Problems|                                          
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adkikani
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 https://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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2017/1 ... -averages/
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chintanpurohit
the answer is
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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S1937
chintanpurohit
the answer is
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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sondenso
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. \(33\frac{1}{3}\)%
C. 40%
D. 50%
E. \(66\frac{2}{3}\)%
Hello Experts.
EMPOWERgmatRichC, VeritasKarishma, IanStewart, Bunuel, chetan2u, ArvindCrackVerbal, GMATGuruNY, AaronPond, GMATinsight, ccooley

In this question prompt, x=40% , y=25%. So, our weighted average should be between 25 and 40.
Can we cross out the choices A,C,D,E because they are not between 25 and 40?
Thanks__
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Asad
sondenso
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. \(33\frac{1}{3}\)%
C. 40%
D. 50%
E. \(66\frac{2}{3}\)%
Hello Experts.
EMPOWERgmatRichC, VeritasKarishma, IanStewart, Bunuel, chetan2u, ArvindCrackVerbal, GMATGuruNY, AaronPond, GMATinsight, ccooley

In this question prompt, x=40% , y=25%. So, our weighted average should be between 25 and 40.
Can we cross out the choices A,C,D,E because they are not between 25 and 40?
Thanks__

No, that will not be the correct way.
The weighted average is already given as 30%, so what is in choices is NOT weighted average but the % of X in mix of X+Y.
As a matter of fact, each of the choice will give you weighted average between 25 and 30.

Now, weighted average method will help you in getting the ratio of X to Total, which will be (30-25)/(40-25)=5/15=1/3=33.33%
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Asad
sondenso
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. \(33\frac{1}{3}\)%
C. 40%
D. 50%
E. \(66\frac{2}{3}\)%
Hello Experts.

In this question prompt, x=40% , y=25%. So, our weighted average should be between 25 and 40.
Can we cross out the choices A,C,D,E because they are not between 25 and 40?
Thanks__

Asad

In this question, the required output is not the percentage of ryegrass in mixture else you would have been correct. But what's given here as 30% is not to be caklculated.

Here the question is asking the percentage of X in mixture so it's important to find the ratio of x and y first which has less to do with the percentage range

You need to rather understand that 40 and 25 have made average 30

30 is at distance of 5 from 25
30 is at distance of 10 from 40

i.e. 25% ryegrass solution has double the weightage than 40% ryegrass solution

i.e. x:y = 1:2

i.e. x% = 1/3 = 33.33%

Hope that help! :)
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Asad
sondenso
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. \(33\frac{1}{3}\)%
C. 40%
D. 50%
E. \(66\frac{2}{3}\)%
Hello Experts.
EMPOWERgmatRichC, VeritasKarishma, IanStewart, Bunuel, chetan2u, ArvindCrackVerbal, GMATGuruNY, AaronPond, GMATinsight, ccooley

In this question prompt, x=40% , y=25%. So, our weighted average should be between 25 and 40.
Can we cross out the choices A,C,D,E because they are not between 25 and 40?
Thanks__

Hi Asad,

In this question, you're mixing together two separate seed mixtures - so unless you have the same amount of each (50% of Mix X and 50% of Mix Y), then you will have MORE than 50% of one of the two Mixes. THAT overall weight is what we're ultimately asked about - meaning that the answer to this question involves more than just the percentages of ryegrass in each mixture. By extension, we cannot simply use the percentage of ryegrass (in this case, 25% and 40%) as a gauge for which answers might be too big or too small. IF the question was rewritten and asked "what percentage of the overall mixture is ryegrass", then the correct answer would have to be between 25% and 40%, but that's not what this prompt asks us for (and as it stands, the prompt TELLS US the overall mixture is 30% ryegrass).

With this prompt, it's essentially really lucky that that thinking got you the correct answer - but it's not mathematically correct.

GMAT assassins aren't born, they're made,
Rich
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sondenso
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. \(33\frac{1}{3}\)%

C. 40%

D. 50%

E. \(66\frac{2}{3}\)%

Answer: Option B

Please check video solution for step by step explanation (TWO OUTSTANDING SOLUTIONS)

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sondenso
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. \(33\frac{1}{3}\)%

C. 40%

D. 50%

E. \(66\frac{2}{3}\)%
We are provided information about rye grass, hence let's focus on the same.

\(\frac{0.40X + 0.25Y}{X + Y} = 0.30\)

\(40X + 25Y = 30X + 30Y\)

\(10X = 5Y\)

\(2X = Y\)

\(\frac{X}{Y} = \frac{1}{2}\)

\(\frac{X}{X+Y} = \frac{1}{1 + 2}\)

\(\frac{X}{X+Y} = \frac{1}{3}\) or \(33\frac{1}{3}\)%

Ans. B
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sondenso
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. \(33\frac{1}{3}\)%

C. 40%

D. 50%

E. \(66\frac{2}{3}\)%

https://gmatclub.com/forum/tips-and-tri ... l#p1218895 This should do
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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%

Answer: B

JeffTargetTestPrep

Can I create one more expression as this below?
X + Y = 100
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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et 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 mi
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Video solution from TGS - The GMAT® Strategy:

Want to reach your dream GMAT® score in half the normal time? Free resources to show you how: http://linktr.ee/thegmatstrategy
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sondenso
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. \(33\frac{1}{3}\)%

C. 40%

D. 50%

E. \(66\frac{2}{3}\)%

Came across the simpler- Alligation method- see attached image for method.
Reference link I found on youtube- https://www.youtube.com/watch?v=C81cWy_GQQM

Do check it out for mixtures type qs
Attachments

Alligation.png
Alligation.png [ 1.05 MiB | Viewed 2435 times ]

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Short video solution here, using the Teeter Totter Timesaver Method (2:06):




Teeter Totter Method Steps:

1) Draw the Endpoints and Weighted Average numbers on the Teeter Totter Diagram (the weighted average represents the final mixture, and goes above the triangle-shaped fulcrum)

2) Draw the weights on each side. The LARGER weight goes on the side with the SHORTER distance from the fulcrum

3) Draw the distances from the Endpoints to the Weighted Average

4) Solve the equation: (W2)/(W1) =(D1)/(D2)
(the weights and distances are in an INVERSE RATIO)­


Teeter Totter Basic Examples Playlist: https://www.youtube.com/playlist?list=PL2exXfCUscn8Hvafet5-IPH1eNNLSjQBP
Attachments

2023-12-29 23_15_27-No name - 29 December 2023 (6).mp4 - VLC media player.png
2023-12-29 23_15_27-No name - 29 December 2023 (6).mp4 - VLC media player.png [ 3.54 MiB | Viewed 1640 times ]

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