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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 09 Jan 2018, 05:47
chetan2u wrote:
dave13 wrote:
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%


Hello everyone :)

ok here is my "correct" soloution

i used formula of weighted average:

\(\frac{W1}{W2}\) =\(\frac{Avg2 - Avg}{Avg-Avg1}\)

X mixture = .40ryegrass + .60bluegrass
Y mixture = .25ryegrass + .75fescue

combined ryegrass .40ryegrass + .25ryegrass = 30%

so :-) i asked myself a question which one X.40 or Y.25 has more amount of ryegrass ? so the one that is closer to combined Average .30, has more amount of rygrass - hence Y.25

So, W1 = Y.25 and W2 =X.40 --->\(\frac{W1}{W2}\) = \(\frac{40 - 30}{30-25}\) =\(\frac{10}{5}\) = 2

So where am i wrong ? i did exactly, as Bunuel explained here :) https://gmatclub.com/forum/weighted-ave ... 06999.html

perhaps you friends can help :? :-) chetan2u , niks18


Hi..

you have gone wrong in taking base of weighted average..
you are looking for % of entire mixture and that will be 40-25..
x will be less as 30 is closer to y(25) so x will be 30-25..
ans \(\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33\)%

Hope it helps


Many thanks chetan2u for reply and explanation!

one question: can you please elaborate on this ---> " you are looking for % of entire mixture and that will be 40-25 " yes % of the entire mixture why are you subracting 25 from 40 :?

shouldnt it be 40%+25% = 30 :?

i followed the paatern as explained here in this question (extrac:ted from the link above)

“Mixture A is 20 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 15 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?
Can you say anything about the relative value of V1 and V2? Can you say which one will be greater? We discussed in the previous post that when we find the weighted average of two quantities, the average is closer to the one which has higher weight because that quantity ‘pulls’ the average toward itself. So if the new mixture contains 30% alcohol, which of the two mixtures – A (20% alcohol) or B (50% alcohol), will have higher volume? Obviously A because it pulled the average toward itself. The average, 30% alcohol, is closer to 20% than to 50%. Now, let’s calculate exactly how much of the 15 gallons was mixture A and how much was mixture B.
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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New post 09 Jan 2018, 05:59
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dave13 wrote:
chetan2u wrote:
Hi..

you have gone wrong in taking base of weighted average..
you are looking for % of entire mixture and that will be 40-25..
x will be less as 30 is closer to y(25) so x will be 30-25..
ans \(\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33\)%

Hope it helps


Many thanks chetan2u for reply and explanation!

one question: can you please elaborate on this ---> " you are looking for % of entire mixture and that will be 40-25 " yes % of the entire mixture why are you subracting 25 from 40 :?

shouldnt it be 40%+25% = 30 :?

i followed the paatern as explained here in this question (extrac:ted from the link above)

“Mixture A is 20 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 15 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?
Can you say anything about the relative value of V1 and V2? Can you say which one will be greater? We discussed in the previous post that when we find the weighted average of two quantities, the average is closer to the one which has higher weight because that quantity ‘pulls’ the average toward itself. So if the new mixture contains 30% alcohol, which of the two mixtures – A (20% alcohol) or B (50% alcohol), will have higher volume? Obviously A because it pulled the average toward itself. The average, 30% alcohol, is closer to 20% than to 50%. Now, let’s calculate exactly how much of the 15 gallons was mixture A and how much was mixture B.


Hi..
here we have common item between the two mix as ryegrass..
it is 40% in X and 25% in Y... the mix gets it to 30%..
1) first point is if both were equal, the average would be 40+25/2=33.5%....... But here it is 30% and 30% is closer to 25%, so Y is MORE than X..
2) what is the difference between ryegrass separately - it is 40%-25% = 15%..
And WHAT is X doing to Mix - it is getting Y up from 25% to 30%, so it is COVERING the 5% of the earlier difference 15% and that is WHY its ratio is 5/15..
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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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

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New post 10 Jan 2018, 00:41
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Using Alligation(to recognise is the difficult part),
we get 33.3%. B
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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New post 24 Jan 2018, 10:01
chetan2u wrote:
dave13 wrote:
chetan2u wrote:
Hi..

you have gone wrong in taking base of weighted average..
you are looking for % of entire mixture and that will be 40-25..
x will be less as 30 is closer to y(25) so x will be 30-25..
ans \(\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33\)%

Hope it helps


Many thanks chetan2u for reply and explanation!

one question: can you please elaborate on this ---> " you are looking for % of entire mixture and that will be 40-25 " yes % of the entire mixture why are you subracting 25 from 40 :?

shouldnt it be 40%+25% = 30 :?

i followed the paatern as explained here in this question (extrac:ted from the link above)

“Mixture A is 20 percent alcohol, and mixture B is 50 percent alcohol. If the two are poured together to create a 15 gallon mixture that contains 30 percent alcohol, approximately how many gallons of mixture A are in the mixture?
Can you say anything about the relative value of V1 and V2? Can you say which one will be greater? We discussed in the previous post that when we find the weighted average of two quantities, the average is closer to the one which has higher weight because that quantity ‘pulls’ the average toward itself. So if the new mixture contains 30% alcohol, which of the two mixtures – A (20% alcohol) or B (50% alcohol), will have higher volume? Obviously A because it pulled the average toward itself. The average, 30% alcohol, is closer to 20% than to 50%. Now, let’s calculate exactly how much of the 15 gallons was mixture A and how much was mixture B.


Hi..
here we have common item between the two mix as ryegrass..
it is 40% in X and 25% in Y... the mix gets it to 30%..
1) first point is if both were equal, the average would be 40+25/2=33.5%....... But here it is 30% and 30% is closer to 25%, so Y is MORE than X..
2) what is the difference between ryegrass separately - it is 40%-25% = 15%..
And WHAT is X doing to Mix - it is getting Y up from 25% to 30%, so it is COVERING the 5% of the earlier difference 15% and that is WHY its ratio is 5/15..



Hi niks18, i am thinking about Chetan`s explanation and solution :)

if i follow strictly weighted average formula i get this --- > X/Y = (25 - 30)/(30 - 40) = 1/2 but not 1/3

but Chetan did this \(\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33\)

perhaps you could explain the difference :? kinda confused, is he using some other weighted average formula ?

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

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dave13 wrote:
Hi niks18, i am thinking about Chetan`s explanation and solution :)

if i follow strictly weighted average formula i get this --- > X/Y = (25 - 30)/(30 - 40) = 1/2 but not 1/3

but Chetan did this \(\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33\)

perhaps you could explain the difference :? kinda confused, is he using some other weighted average formula ?

thank you :)


Hi dave13

the value that you got in the highlighted portion is correct.

so we have \(\frac{X}{Y}=\frac{1}{2}\)

Hence % of X in the mixture will be \(= \frac{1}{(1+2)}*100=\frac{1}{3}*100=33.33\)%

The question is about % of X in the mixture and not about ratio of X & Y.
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Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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New post 24 Jan 2018, 11:22
niks18 wrote:
dave13 wrote:
Hi niks18, i am thinking about Chetan`s explanation and solution :)

if i follow strictly weighted average formula i get this --- > X/Y = (25 - 30)/(30 - 40) = 1/2 but not 1/3

but Chetan did this \(\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33\)

perhaps you could explain the difference :? kinda confused, is he using some other weighted average formula ?

thank you :)


Hi dave13

the value that you got in the highlighted portion is correct.

so we have \(\frac{X}{Y}=\frac{1}{2}\)

Hence % of X in the mixture will be \(= \frac{1}{(1+2)}*100=\frac{1}{3}*100=33.33\)%

The question is about % of X in the mixture and not about ratio of X & Y.



niks18 thank you for quick reply :) i have one silly question :) doesnt numerator show how many "items" we have of the total and denominator shows how many we have in total ? :)

how from this 1/2 you got this \(\frac{1}{(1+2)}\)--- arent there maximum 2 "items" :?
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Re: Seed mixture X is 40 percent ryegrass and 60 percent [#permalink]

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New post 24 Jan 2018, 18:31
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dave13 wrote:
niks18 wrote:
dave13 wrote:
Hi niks18, i am thinking about Chetan`s explanation and solution :)

if i follow strictly weighted average formula i get this --- > X/Y = (25 - 30)/(30 - 40) = 1/2 but not 1/3

but Chetan did this \(\frac{30-25}{40-25}=\frac{5}{15} = 1/3 = 33.33\)

perhaps you could explain the difference :? kinda confused, is he using some other weighted average formula ?

thank you :)


Hi dave13

the value that you got in the highlighted portion is correct.

so we have \(\frac{X}{Y}=\frac{1}{2}\)

Hence % of X in the mixture will be \(= \frac{1}{(1+2)}*100=\frac{1}{3}*100=33.33\)%

The question is about % of X in the mixture and not about ratio of X & Y.



niks18 thank you for quick reply :) i have one silly question :) doesnt numerator show how many "items" we have of the total and denominator shows how many we have in total ? :)

how from this 1/2 you got this \(\frac{1}{(1+2)}\)--- arent there maximum 2 "items" :?


Hi dave13

You got the ratio of x & y so here numerator will be quantity of x that is 1 part and denominator will be y that is 2 parts. So tota quantity of mixture x+y=1+2=3

Hence % of x=1/3

This is a simple ratio problem.

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Re: Seed mixture X is 40 percent ryegrass and 60 percent   [#permalink] 24 Jan 2018, 18:31

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