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A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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30 Sep 2016, 05:20
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A new beverage, which is a mixture of grape juice and apple juice, is being tested. The test container of 800 milliliters is 15% grape juice. If the response to the beverage causes the maker to increase the percentage of grape juice in the container to 20%, how much grape juice will be added? A) 30 B) 40 C) 50 D) 120 E) 160
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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30 Sep 2016, 05:59
Don't know how the answer is C) 50.
15%  the volume is 120
if it's increased to 20% of the total volume, we're looking at 160
The difference is 40. IMO, this should be the answer.
Can someone help me out?



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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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30 Sep 2016, 06:22
I also got B for the same reasons. is the OA mentioned correct?
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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duahsolo wrote: A new beverage, which is a mixture of grape juice and apple juice, is being tested. The test container of 800 milliliters is 15% grape juice. If the response to the beverage causes the maker to increase the percentage of grape juice in the container to 20%, how much grape juice will be added?
A) 30 B) 40 C) 50 D) 120 E) 160 Hi vp101 & abhimahna, I can confirm that the OA is correct. Now to the question: Let's say the amount of grape juice to be added to increase the percentage of grape juice in the container to 20% is x ORIGINAL %age of grape juice in the container = 120/800 = 15% NEW %age of grape juice in the container = (120 + x) / ( 800 + x) = 20% > 600 + 5x = 800 + x > x = 50 Answer is C (50)
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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30 Sep 2016, 06:31
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duahsolo wrote: duahsolo wrote: A new beverage, which is a mixture of grape juice and apple juice, is being tested. The test container of 800 milliliters is 15% grape juice. If the response to the beverage causes the maker to increase the percentage of grape juice in the container to 20%, how much grape juice will be added?
A) 30 B) 40 C) 50 D) 120 E) 160 Hi vp101 & abhimahna, I can confirm that the OA is correct. Now to the question: Let's say the amount of grape juice to be added to increase the percentage of grape juice in the container to 20% is x ORIGINAL %age of grape juice in the container = 120/800 = 15% NEW %age of grape juice in the container = (120 + x) / ( 800 + x) = 20% > 600 + 5x = 800 + x > x = 50 Answer is C (50) Ohh, I missed such a simple point. Anyways, thanks.
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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30 Sep 2016, 08:04
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duahsolo wrote: A new beverage, which is a mixture of grape juice and apple juice, is being tested. The test container of 800 milliliters is 15% grape juice. If the response to the beverage causes the maker to increase the percentage of grape juice in the container to 20%, how much grape juice will be added?
A) 30 B) 40 C) 50 D) 120 E) 160 Quote: The test container of 800 milliliters is 15% grape juice. Grape Juice = 120 ml Water = 680 ml Total Solution = 800 ml Quote: If the response to the beverage causes the maker to increase the percentage of grape juice in the container to 20%, how much grape juice will be added So, \(\frac{(120 + k )}{( 800 + k )}= \frac{20}{100}\) Or, \(\frac{(120 + k )}{( 800 + k )}= \frac{1}{5}\) Or, \(600 + 5k = 800 + k\) Or, \(4k = 200\) Or, \(k = 50\) So, 50 ml of Grape juice , must be added , answer must be (C)
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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02 Oct 2016, 15:48
VeritasPrepKarishma, BunuelIs it possible to do this using weighted avg? I tried after reading your post on 'weighted avg and mixture problems on GMAT' but cant seem to make it work. Is weighted avg used only when the same solution is considered and not two separate like in this case?
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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02 Oct 2016, 16:37
colorblind wrote: VeritasPrepKarishma, BunuelIs it possible to do this using weighted avg? I tried after reading your post on 'weighted avg and mixture problems on GMAT' but cant seem to make it work. Is weighted avg used only when the same solution is considered and not two separate like in this case? VeritasPrepKarishma, BunuelAaahhhhhh I think I got it, please correct me if I am wrong: Initial grape juice percent = 15% = \(\frac{5}{20}\) 100% concentration grape juice = 1 Grape juice concentration in final mixture = 20% = \(\frac{1}{5}\) \(\frac{800}{W2} = (1\frac{1}{5})/(\frac{1}{5} \frac{5}{20})\) W2 = 50.
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A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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02 Oct 2016, 20:58
15% of 800ml means we've got 120 ml of grape juice.
Hmmm... looking at the answer choices we've got some easy well rounded numbers, perfect for plugging in answers. Let's do that.
Starting with B, we add 40ml and have 840ml total, 20% of that is 168ml of grape juice. BUT 120ml of original grape juice plus the 40ml we just added comes out to 160ml (not equal to 168ml). So B doesnt work (it's too low).
Let's try the same thing with answer choice D. We add 120ml of juice to get 920ml total. 20% of that is 184ml, 120ml of original grape juice plus the 120ml we just added comes out to 240ml (not equal to 184ml). So D doesnt work (it's too high).
So the correct answer must be C
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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03 Oct 2016, 03:18
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colorblind wrote: colorblind wrote: VeritasPrepKarishma, BunuelIs it possible to do this using weighted avg? I tried after reading your post on 'weighted avg and mixture problems on GMAT' but cant seem to make it work. Is weighted avg used only when the same solution is considered and not two separate like in this case? VeritasPrepKarishma, BunuelAaahhhhhh I think I got it, please correct me if I am wrong: Initial grape juice percent = 15% = \(\frac{5}{20}\) 100% concentration grape juice = 1 Grape juice concentration in final mixture = 20% = \(\frac{1}{5}\) \(\frac{800}{W2} = (1\frac{1}{5})/(\frac{1}{5} \frac{5}{20})\) W2 = 50. Yes, correct! Though I would use percentages only. They are easier to manipulate compared with fractions. 800/w2 = (100  20)/(20  15) = 16/1 w2 = 50
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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20 Oct 2016, 05:50
Is it possible to do this using weighted avg? I tried after reading your post on 'weighted avg and mixture problems on GMAT' but cant seem to make it work. Is weighted avg used only when the same solution is considered and not two separate like in this case?[/quote]
Aaahhhhhh I think I got it, please correct me if I am wrong:
Initial grape juice percent = 15% = \(\frac{5}{20}\) 100% concentration grape juice = 1 Grape juice concentration in final mixture = 20% = \(\frac{1}{5}\)
\(\frac{800}{W2} = (1\frac{1}{5})/(\frac{1}{5} \frac{5}{20})\)
W2 = 50.[/quote]
Yes, correct! Though I would use percentages only. They are easier to manipulate compared with fractions.
800/w2 = (100  20)/(20  15) = 16/1
w2 = 50[/quote]
Hi VeritasPrepKarishma,
I had tried using the weighted average formula but could not do so. I see that you have performed the following:
Comparing the step 800/w2 = (100  20)/(20  15) = 16/1 to the weighted average formula, i don't seem to figure why we have taken 100 % here.
n1/n2=(A2Aw)/(AwA1). So Aw here is 20, A1 is 15, n1 is 800. How did you take A2 as 100? What is the logic behind it? Can you please explain it to me?



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A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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20 Oct 2016, 15:04
vrgmat wrote: Is it possible to do this using weighted avg? I tried after reading your post on 'weighted avg and mixture problems on GMAT' but cant seem to make it work. Is weighted avg used only when the same solution is considered and not two separate like in this case? Aaahhhhhh I think I got it, please correct me if I am wrong: Initial grape juice percent = 15% = \(\frac{5}{20}\) 100% concentration grape juice = 1 Grape juice concentration in final mixture = 20% = \(\frac{1}{5}\) \(\frac{800}{W2} = (1\frac{1}{5})/(\frac{1}{5} \frac{5}{20})\) W2 = 50.[/quote] Yes, correct! Though I would use percentages only. They are easier to manipulate compared with fractions. 800/w2 = (100  20)/(20  15) = 16/1 w2 = 50[/quote] Hi VeritasPrepKarishma, I had tried using the weighted average formula but could not do so. I see that you have performed the following: Comparing the step 800/w2 = (100  20)/(20  15) = 16/1 to the weighted average formula, i don't seem to figure why we have taken 100 % here. n1/n2=(A2Aw)/(AwA1). So Aw here is 20, A1 is 15, n1 is 800. How did you take A2 as 100? What is the logic behind it? Can you please explain it to me?[/quote] Initial concentration of grape juice is 15% and we are asked to increase the percentage of grape juice to 20%. The only way to do this is by adding a certain amount of 100% grape juice and that is where the 100% comes from.
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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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20 Oct 2016, 22:35
colorblind wrote: vrgmat wrote: Is it possible to do this using weighted avg? I tried after reading your post on 'weighted avg and mixture problems on GMAT' but cant seem to make it work. Is weighted avg used only when the same solution is considered and not two separate like in this case? Aaahhhhhh I think I got it, please correct me if I am wrong: Initial grape juice percent = 15% = \(\frac{5}{20}\) 100% concentration grape juice = 1 Grape juice concentration in final mixture = 20% = \(\frac{1}{5}\) \(\frac{800}{W2} = (1\frac{1}{5})/(\frac{1}{5} \frac{5}{20})\) W2 = 50. Yes, correct! Though I would use percentages only. They are easier to manipulate compared with fractions. 800/w2 = (100  20)/(20  15) = 16/1 w2 = 50[/quote] Hi VeritasPrepKarishma, I had tried using the weighted average formula but could not do so. I see that you have performed the following: Comparing the step 800/w2 = (100  20)/(20  15) = 16/1 to the weighted average formula, i don't seem to figure why we have taken 100 % here. n1/n2=(A2Aw)/(AwA1). So Aw here is 20, A1 is 15, n1 is 800. How did you take A2 as 100? What is the logic behind it? Can you please explain it to me?[/quote] Initial concentration of grape juice is 15% and we are asked to increase the percentage of grape juice to 20%. The only way to do this is by adding a certain amount of 100% grape juice and that is where the 100% comes from.[/quote] Got it. Thank you very much colorblind



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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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10 Mar 2017, 12:53
duahsolo wrote: A new beverage, which is a mixture of grape juice and apple juice, is being tested. The test container of 800 milliliters is 15% grape juice. If the response to the beverage causes the maker to increase the percentage of grape juice in the container to 20%, how much grape juice will be added?
A) 30 B) 40 C) 50 D) 120 E) 160 Hi there, I don't understand why we are using the formula 120+x/800+x to arrive at the answer. The question stem has asked "how much grape juice to be added in the container"? Since the container's capacity is 800 ml we cannot use 800+x as the juice will spill out. @Experts: your thoughts please?



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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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10 Mar 2017, 13:56
First step is to figure out how many ml of grape juice is equal to 15%
15% of 800 is 120 ml.
Now set up your equation
x+120/x+800=20/100
we're adding the grape juice to increase the grape juice concentrate and also adding to the overall mixture. When you solve for X you get 50



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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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10 Mar 2017, 16:31
duahsolo wrote: A new beverage, which is a mixture of grape juice and apple juice, is being tested. The test container of 800 milliliters is 15% grape juice. If the response to the beverage causes the maker to increase the percentage of grape juice in the container to 20%, how much grape juice will be added?
A) 30 B) 40 C) 50 D) 120 E) 160 let x=ml of grape juice to be added .15*800+x=.2(800+x) x=50 ml C



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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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11 Mar 2017, 08:18
this is poorly worded. if the test container is of 800ml, why should i assume that it can hold 850ml??



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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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11 Mar 2017, 10:34
rt9 wrote: this is poorly worded. if the test container is of 800ml, why should i assume that it can hold 850ml?? The exact same assumption stumped me too. I interpreted the container to be of 800 ml Volume. Now to me, "response", meant that someone would suck out some volume of liquid from the 800 ml filled container resulting in reduction of the liquid Now one has to add some X ml of Grape juice to make it back to 800 ml mixture Hence I computed (120 + X)/800= 0.2 resulting in X= 40 May be I interpreted it very incorrectly KM



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Re: A new beverage, which is a mixture of grape juice and apple juice, is [#permalink]
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04 Sep 2017, 20:10
I will better use options here, 15% of 800 = 120 So adding grape juice from options Starting with option C: Option C: Total grape = 120+50= 170 Total Juice = 800+50=850 20% of 850 = 170 Thus option C
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