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A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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17 Dec 2009, 15:48
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A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A? A) 40% B) 45% C) 50 % D) 60 % E) 55 %
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Re: Percent Mix Problem [#permalink]
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17 Dec 2009, 16:21
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Another alligation method:
Set up a table: Millet Sunflower Brand A 40% 60% Brand B 65% 35%
BAverageA (from mean) (from mean) 15 50 10
So A/A+B = 15/25 = 60%



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Re: Percent Mix Problem [#permalink]
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17 Dec 2009, 20:58
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this can be done easily with the tricks of mix and alligation ans is 60% 40________________________65 ____________50____________ 6550=15______:_______ 5040=10 so ratio is 15:10 % is 15/25 *100=60
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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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19 May 2014, 22:17
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Another method to solve this question using equations: 1 kg of Brand A will have 400g millet and 600g sunflower. 1 kg of Brand B will gave 650g millet and 350g sunflower. Suppose,x kg of Brand A and y kg of Brand B are in the mixture. Then total weight of millet = 400x+650y g Total weight of mixture = 1000x + 1000y g Given that millet is 500% hence (400x+650y)/(1000x + 1000y) = 1/2 or, 800x + 1300y = 1000x+1000y or 300y=200x or, x/y = 3/2 % of Brand A = (3/(3+2))*100 =60% Hence answer is D
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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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02 Jun 2014, 18:43
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Took me 1min.
Only take the Millet percentages since that's the goal mix you´re trying to find.
0.40x + 0.65(1x) = 0.50 0.40x+ 0.65  0.65x = 0.50 0.25x = 0.15 x = 0.60
Hope it helps!



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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02 Jun 2014, 20:47
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umaa wrote: A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A?
A) 40%
B) 45%
C) 50 %
D) 60 %
E) 55 % Use the scale method: 40% millet and 65% millet are mixed together to get 50% millet. w1/w2 = (65  50)/(50  40) = 15/10 = 3/2 So brand A is 3/5 i.e. 60% of the mix. Answer (D) For more on this method, check: http://www.veritasprep.com/blog/2011/03 ... averages/
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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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01 Jul 2014, 08:55
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why r we taking 15/10 and not 10/15 when 10 denotes brand A part .?



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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02 Jul 2014, 22:31
kunaljain1701 wrote: why r we taking 15/10 and not 10/15 when 10 denotes brand A part .? It's actually 15 parts of brand A and 10 parts of brand B. So fraction of brand A is 15/25. The w1 and w2 in the formula are the weights of brand A and brand B. They do not represent the fraction of total but only the parts of brand A and brand B. Check out the link I have given above. It discusses weights in more detail.
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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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23 Oct 2014, 04:55
Bumping up and asking for a different explanation. So, according to MGMAT's explanation, they used a table method to solve this question  which is not very helpful. Could someone explain in a simpler way to solve this using the table method? Also, isn't there "one ring to rule them all" for this type of questions? p.s.: Sadly,I can't attach a screen grab of their explanation to this reply



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A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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23 Oct 2014, 06:52
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Blackbox wrote: Bumping up and asking for a different explanation. So, according to MGMAT's explanation, they used a table method to solve this question  which is not very helpful. Could someone explain in a simpler way to solve this using the table method? Also, isn't there "one ring to rule them all" for this type of questions? p.s.: Sadly,I can't attach a screen grab of their explanation to this reply A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A? A) 40% B) 45% C) 50 % D) 60 % E) 55 % Yes there is a simple method : Consider the following method Brand A : 40% millet and 60% sunflower Brand B : 65% millet and 35% safflower Mix : 50% millet Here the weighted average is 50%, Now Brand A has 40% millet, which is 10% less than the weighted average of mix =  0.10 A  I Similarly, Brand B has 65 % millet, which is 15 % more than the weighted average of mix = + 0.15 B  II Now, both Brand A and Brand B are combined to give a 50% mix containing millet, so equate I and II implies, 0.10 A = 0.15 B Therefore A/B = 0.15/0.10 = 3/2 A : B : (A + B) = 3 : 2 : (3+2) = 3 : 2 : 5 We have to find, percent of the mix is Brand A i.e. A : (A + B) = 3 : 5 = (3 / 5) * 100 = 60 % Here is a pictorial representation : Brand A= 40%10% or 0.10 below average, A times Total below =  0.10 A  Average = 50% or 0.50 Brand B = 65 % 15% or 0.15 above average, B times Total above = + 0.15 B Since the amount below the average has to equal the average above the average; therefore, 0.10 A = 0.15 BA/B = 3/2
A:B: Total = 3:2:5 Therefore A/Total = 3:5 = 60 %



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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23 Oct 2014, 10:26
Thanks for the explanation, Ashish.
1. Does this approach work with all sorts of problems with percents? 2. How did you come up with 50% as a weighted average? Is it because it is given "If a customer purchases a mix of the two types of birdseed that is 50% millet"?



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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23 Oct 2014, 21:08
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Blackbox wrote: Thanks for the explanation, Ashish.
1. Does this approach work with all sorts of problems with percents? 2. How did you come up with 50% as a weighted average? Is it because it is given "If a customer purchases a mix of the two types of birdseed that is 50% millet"? This approach works for mixture problems. Its better to convert the information into ratios. It causes less confusion. This is a weighted avg problem of sorts, because you got 2 Brands and the final mix contains some part of Brand and some part of Brand B. This is evident from the fact that Brand A has 40 % millet, Brand B has 65 % millet and Mix has 50 % millet 40% Millet in Brand A 50% Millet in Mix  65% Millet in Brand B. As you see, Brand A and Brand B combine to form the Mix. Had the Quantity of Brand A and Quantity of Brand B been Equal then the Mix would have had 52.5% Millet. But this is not the case here. We are given that the Mix has 50% millet i.e. the Quantity of Brand A in mix is greater than Quantity of Brand B in mix since 50% is closer to 40% than it is to 65%. So, this situation can only arise if and only if this was a weighted average problem. In case you need the Weighted Avg formula, Weighted Avg = [(Quantity of A*Quality of A) + (Quantity of B*Quality of B)] / ( Quantity of A + Quantity of B )



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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23 Oct 2014, 21:11
Ashishmathew01081987 wrote: Blackbox wrote: Thanks for the explanation, Ashish.
1. Does this approach work with all sorts of problems with percents? 2. How did you come up with 50% as a weighted average? Is it because it is given "If a customer purchases a mix of the two types of birdseed that is 50% millet"? This approach works for mixture problems. Its better to convert the information into ratios. It causes less confusion. This is a weighted avg problem of sorts, because you got 2 Brands and the final mix contains some part of Brand and some part of Brand B. This is evident from the fact that Brand A has 40 % millet, Brand B has 65 % millet and Mix has 50 % millet 40% Millet in Brand A 50% Millet in Mix  65% Millet in Brand B. As you see, Brand A and Brand B combine to form the Mix. Had the Quantity of Brand A and Quantity of Brand B been Equal then the Mix would have had 52.5% Millet. But this is not the case here. We are given that the Mix has 50% millet i.e. the Quantity of Brand A in mix is greater than Quantity of Brand B in mix since 50% is closer to 40% than it is to 65%. So, this situation can only arise if and only if this was a weighted average problem. In case you need the Weighted Avg formula, Weighted Avg = [(Quantity of A*Quality of A) + (Quantity of B*Quality of B)] / ( Quantity of A + Quantity of B )Also, keep an eye on the value that is changing. Here, the value is that of the proportion of Millet. Sunflower and safflower values are of no use.



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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25 Oct 2014, 00:24
hey guys , m satisified with the above explanations but let me give u an easy one which is undersble by a newbie. this is very a simple approach u can also find same question in og13(i guess in the last questions) consider brand A40(millet) and 60(other one) n brand B 0.65x(millet) and 0.35x(other one) acc to question the eq is 40+ 0.65x=0.5(x+100) u get x=200/3 now percentage =brand A/mixture 100/(200/3) +100 =3/5 which is 60% simple D h



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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26 Oct 2014, 06:39
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hsbinfy wrote: hey guys , m satisified with the above explanations but let me give u an easy one which is undersble by a newbie. this is very a simple approach u can also find same question in og13(i guess in the last questions) consider brand A40(millet) and 60(other one) n brand B 0.65x(millet) and 0.35x(other one) acc to question the eq is 40+ 0.65x=0.5(x+100) u get x=200/3 now percentage =brand A/mixture 100/(200/3) +100 =3/5 which is 60% simple D h Yup. That's one other explanation I read on MGMAT. Thanks for bringing it up, HSBINFY.



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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20 Apr 2015, 11:14
Could anyone explain why this solution is incorrect;
0.4x+0.65y=0.5(x+y) 0.15y=0.1x > Total weight is 0.25 x= 40% of total.
For reference in this problem this approach works.
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%
0.4x+0.25y = 0.3(x+y) 0.4x0.3x = 0.3y  0.25y 0.1x=0.05y or 2x=y So Y=33 1/3 %



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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20 Apr 2015, 18:58
Hi JeroenReunis, Your calculations are correct; you make a mistake in your end deduction though. When you get to this calculation.... .15Y = .1X You have a RATIO of X to Y. Here's what you need to do with this ratio to get to the correct answer: Let's multiply both sides by 100 to get rid of the decimal.... 15Y = 10X From here, you can do work in a couple of different ways. I'm going to continue to do "ratio math".... 15Y = 10X 15/10 = X/Y 3/2 = X/Y This means that for every 3 "units" of X, we have 2 "units" of Y. So, for every 5 total units, 3/5 are X and 2/5 are Y. The question asks for the percent of the mixture that is X... 3/5 = 60% Final Answer: GMAT assassins aren't born, they're made, Rich
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A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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21 Apr 2015, 02:43
EMPOWERgmatRichC wrote: Hi JeroenReunis, Your calculations are correct; you make a mistake in your end deduction though. When you get to this calculation.... .15Y = .1X You have a RATIO of X to Y. Here's what you need to do with this ratio to get to the correct answer: Let's multiply both sides by 100 to get rid of the decimal.... 15Y = 10X From here, you can do work in a couple of different ways. I'm going to continue to do "ratio math".... 15Y = 10X 15/10 = X/Y 3/2 = X/Y This means that for every 3 "units" of X, we have 2 "units" of Y. So, for every 5 total units, 3/5 are X and 2/5 are Y. The question asks for the percent of the mixture that is X... 3/5 = 60% Final Answer: GMAT assassins aren't born, they're made, Rich Hello Rich, First of all many thanks for your reply, this question can't get out of my mind. However I still do not get it. Since 10x = 15y I would assume that for every 1X there are 1.5Y. So for every 2X there are 3Y. (Ratio is 2:3) So in total there are 5 units, X= 2/5 of Total which is 40%. Even after I put down 10 paperclips (=X) and 15 coins(=Y) on my desk, i still see it as there are 1.5Y for every 1X. Another example if the ratio of cows to pigs is 2 to 4 than 1C=2P right? How is this any different? Could you please try to explain to me where my thinking goes wrong? Just don't get it



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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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umaa wrote: A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sunflower, and Brand B, which is 65% millet and 35% safflower. If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A?
A) 40%
B) 45%
C) 50 %
D) 60 %
E) 55 %
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Re: A feed store sells two varieties of birdseed: Brand A, which [#permalink]
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21 Apr 2015, 18:48
Hi JeroenReunis, The given equation can be rewritten as a ratio, but your interpretation of it is not correct.... 10X = 15Y means "10 times X" equals "15 times Y" This does NOT mean "for every 10X there are 15Y" You can prove this by TESTing VALUES. 10X = 15Y IF.... X = 3 then Y = 2 THIS is the ratio that actually exists: X:Y = 3 to 2 Any additional sets of values that you plug in for X and Y will confirm this ratio. GMAT assassins aren't born, they're made, Rich
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