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A new beverage, which is a mixture of grape juice and apple juice, is

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A new beverage, which is a mixture of grape juice and apple juice, is  [#permalink]

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New post 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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New post 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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New post 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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New post 30 Sep 2016, 06:27
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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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New post 30 Sep 2016, 08:04
1
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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New post 02 Oct 2016, 15:48
VeritasPrepKarishma, Bunuel

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?
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Re: A new beverage, which is a mixture of grape juice and apple juice, is  [#permalink]

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New post 02 Oct 2016, 16:37
colorblind wrote:
VeritasPrepKarishma, Bunuel

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?


VeritasPrepKarishma, Bunuel

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.
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Re: A new beverage, which is a mixture of grape juice and apple juice, is  [#permalink]

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New post 03 Oct 2016, 03:18
1
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colorblind wrote:
colorblind wrote:
VeritasPrepKarishma, Bunuel

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?


VeritasPrepKarishma, Bunuel

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
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Re: A new beverage, which is a mixture of grape juice and apple juice, is  [#permalink]

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New post 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=(A2-Aw)/(Aw-A1). 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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New post 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=(A2-Aw)/(Aw-A1). 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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New post 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=(A2-Aw)/(Aw-A1). 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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New post 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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New post 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

:)
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Re: A new beverage, which is a mixture of grape juice and apple juice, is  [#permalink]

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New post 19 Mar 2017, 13:55
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2
In the original Test Container of 800 milliliters, grape juice is 15%, i.e. 120 milliliters

To make the grape juice % to 20%, let the amount of grape juice added = x milliliters

(120+x)/(800+x) = 0.2

Solving for X gives as 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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New post 19 Mar 2017, 18:15
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Bunuel 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



Initial volume of grape juice in beverage 15% of 800 = 120ML
let X amount be added then
this new volume will represent 20% of final volume
120+x = 20% of (800+x)
solving X =50 ML

Ans 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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New post 20 Mar 2017, 00:20
Bunuel 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 the grape juice added be x
apple juice in initial mixture = 0.85*800
apple juice in final mixture = 0.8(800 +x)
since the apple juice in both the mixture is same, therefore
0.85*800 = 0.8(800+ x)
0.05*800 = 0.8x
or x= 0.05*800/0.8 = 50ml

Option 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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New post 24 Mar 2017, 17:30
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Bunuel 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


800 ml, 15% grape juice = 120ml grape & 85% apple = 680.

Now 680 must be 80% of new mixture so find value that 680ml is 80% of. so (680/8)*10 = 850 hence 50 more.
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Re: A new beverage, which is a mixture of grape juice and apple juice, is  [#permalink]

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New post 31 May 2018, 15:09
Bunuel 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


Initially there is 800 x 0.15 = 120 ml of grape juice and thus 680 ml of apples juice. We can let g = the amount of grape juice to be added and create the equation:

(120 + g)/(800 + g) = 1/5

5(120 + g) = 800 + g

600 + 5g = 800 + g

4g = 200

g = 50

Alternate Solution:

To 800 ml of 15% grape juice, we add x ml of 100% grape juice to obtain (800 + x) ml of 20% grape juice. Converting the percentages to decimals, we can create the equation:

800(0.15) + x(1.0) = (800 + x)(0.20)

120 + x = 160 + 0.20x

0.8x = 40

8x = 400

x = 50

Answer: 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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New post 23 Jun 2018, 07:05
Bunuel 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(100 % grapes juice mixture) is required to be added to make the final % of grape juice to be 20 %

therefore
800/x = (100 - 20)/(20-15)

x = 800/16 = 50 %
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Re: A new beverage, which is a mixture of grape juice and apple juice, is &nbs [#permalink] 23 Jun 2018, 07:05
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