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

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

Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]
03 Jun 2014, 20:45

Expert's post

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%

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% _________________

Re: 223 Percentage [#permalink]
03 Jun 2014, 21:47

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%)

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]
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:

Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]
09 Nov 2015, 01:03

Expert's post

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%)

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

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