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Re: Seed mixture X is 40 percent ryegrass and 60 percent
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09 Jan 2018, 06:47
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: \(\frac{W1}{W2}\) =\(\frac{Avg2  Avg}{AvgAvg1}\) 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 >\(\frac{W1}{W2}\) = \(\frac{40  30}{3025}\) =\(\frac{10}{5}\) = 2 So where am i wrong ? i did exactly, as Bunuel explained here :) https://gmatclub.com/forum/weightedave ... 06999.htmlperhaps you friends can help :? :) chetan2u , niks18Hi.. you have gone wrong in taking base of weighted average.. you are looking for % of entire mixture and that will be 4025.. x will be less as 30 is closer to y(25) so x will be 3025.. ans \(\frac{3025}{4025}=\frac{5}{15} = 1/3 = 33.33\)% 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 4025 " 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
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09 Jan 2018, 06:59
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 4025.. x will be less as 30 is closer to y(25) so x will be 3025.. ans \(\frac{3025}{4025}=\frac{5}{15} = 1/3 = 33.33\)%
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 4025 " 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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Re: Seed mixture X is 40 percent ryegrass and 60 percent
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10 Jan 2018, 01:41
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
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24 Jan 2018, 11:01
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 4025.. x will be less as 30 is closer to y(25) so x will be 3025.. ans \(\frac{3025}{4025}=\frac{5}{15} = 1/3 = 33.33\)%
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 4025 " 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 \(\frac{3025}{4025}=\frac{5}{15} = 1/3 = 33.33\) 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
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24 Jan 2018, 11:28
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 \(\frac{3025}{4025}=\frac{5}{15} = 1/3 = 33.33\) perhaps you could explain the difference kinda confused, is he using some other weighted average formula ? thank you Hi dave13the 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
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24 Jan 2018, 12:22
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 \(\frac{3025}{4025}=\frac{5}{15} = 1/3 = 33.33\) perhaps you could explain the difference kinda confused, is he using some other weighted average formula ? thank you Hi dave13the 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
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24 Jan 2018, 19:31
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 \(\frac{3025}{4025}=\frac{5}{15} = 1/3 = 33.33\) perhaps you could explain the difference kinda confused, is he using some other weighted average formula ? thank you Hi dave13the 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 dave13You 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. Posted from my mobile device



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Seed mixture X is 40 percent ryegrass and 60 percent
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25 Mar 2018, 08:03
chintanpurohit wrote: the answer is E.66 2/3%
solve for 25x + 40 [1x] = 30
so, x = 2/3 = 66 2/3% You have actually found the weight of Y in the resultant mixture. For the correct answer, you have to do one step further: Weight of X in the mixture= 1 (2/3) = 33.33% Answer is: B



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Re: Seed mixture X is 40 percent ryegrass and 60 percent
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28 Jul 2018, 07:48
chintanpurohit wrote: the answer is E.66 2/3%
solve for 25x + 40 [1x] = 30
so, x = 2/3 = 66 2/3% How did u get 1x?9



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Seed mixture X is 40 percent ryegrass and 60 percent
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04 Aug 2018, 20:24
S1937 wrote: chintanpurohit wrote: the answer is E.66 2/3%
solve for 25x + 40 [1x] = 30
so, x = 2/3 = 66 2/3% How did u get 1x?9 Hi S1937, It turns out that [1x] is an error, if x is meant to represent the fraction of the final mixture made up by X. Instead, [1x] should be the fraction of the final mixture that is made up by Y, and the equation should read: 25[1x] + 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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