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1. x and y should have been mixed in the ratio, 30-25 / 40-30 = 1/2 to get the resulting mixture 2. So the weight of x in the resulting mixture is 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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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%

The answer is B

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

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

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11 Nov 2013, 20:01

1

This post received KUDOS

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%

Its very simple ..

30/100(x+y)= 40/100(x) + 25/100(y) after solving we'll get, y=2x ... and we know x + y = 100(Total mixture)

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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Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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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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.

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

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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:

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

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 %

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:

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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Jeffery Miller Head of GMAT Instruction

GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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.