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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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New post 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 [#permalink]

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New post 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!

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

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New post 31 Jan 2017, 17:16
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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New post 05 Mar 2017, 04:02
GMAT TIGER wrote:
sondenso wrote:
240. 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%


B. x = weight of x
40 x + 25 (1-x) = 30
solve for x, x = 33.33%



40 x + 25 (1-x) = 30
Please explain this
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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New post 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 percent ryegrass and 60 percent [#permalink]

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New post 04 Sep 2017, 02:48
Bunuel, abhimahna, and experts

When should we assume 100 in a question because I assumed both mixtures to be 100 but did not get the answer?
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Re: Seed mixture X is 40% ryegrass and 60% bluegrass by weight; seed mixtu [#permalink]

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New post 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%

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

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New post 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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New post 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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New post 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] 31 Oct 2017, 04:46

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