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A container holds 4 quarts of alcohol and 4 quarts of water. How many

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Veritas Prep GMAT Instructor
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 05 Oct 2017, 02:10
sameerkamath21 wrote:
VeritasPrepKarishma

Thank you so much Karishma. Just one doubt. The A2 and A1 that you used in your blog post about the same topic keep changing as per the question right? And what will w1/w2 be in this case?As in acc to this question?



I am not sure I understand what your question is. Do you mean to ask how we decide on how to allocate subscripts?

This post shows one way: https://gmatclub.com/forum/a-container- ... l#p1935591

Alternatively, pure water can be A1, w1 and the 50-50 mix can be A2, w2. The answer obtained will be the same.

Again, you worked with concentration of water here. You could work with the concentration of alcohol too (pure water having 0 alcohol).
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 06 Oct 2017, 10:56
shrive555 wrote:
A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

A. 4/3
B. 5/3
C. 7/3
D. 8/3
E. 10/3


We can create the following equation in which n = the added water:

4/(4 + n) = 3/5

20 = 12 + 3n

8 = 3n

8/3 = n

Answer: D
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A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 07 Jan 2018, 18:50
shrive555 wrote:
A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

A. 4/3
B. 5/3
C. 7/3
D. 8/3
E. 10/3


let w=quarts of water to be added
4+w=(5/8)(8+w)
w=8/3 quarts
D
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A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 25 Jan 2018, 13:21
VeritasPrepKarishma wrote:
sameerkamath21 wrote:
VeritasPrepKarishma

Hello Karishma. I recently read your blog posts on weighted averages and found the w1/w2 method and the scale method to be very useful, fast and accurate. So i've just started solving mixture problems using your techniques as much as possible. Will take sometime for me to master it. Could you please explain the how you can solve this sum using your method? Thanks



Right now the concentration of water in the mix is 4/8. We want to add pure water (8/8) to it and make the concentration of water in the final solution (5/8).

w1/w2 = (8/8 - 5/8)/(5/8 - 4/8) = 3/1

So for every 3 parts of the original mix, we should add 1 part of pure water.
The original mix is 8 quarts. So we should add 8/3 quarts of pure water.

Answer (D)


Hi VeritasPrepKarishma,

thanks for the link you shared in the previous post https://www.veritasprep.com/blog/2017/1 ... -averages/

that confusion regarding nnomenclature is clear :) now after reading the explanation which is followed by an example i have a question:

regarding this : w1/w2 = (8/8 - 5/8)/(5/8 - 4/8) = 3/1

where does it say that we need to add 8/8 pure water, it says 3 parts of alcohol and only 5 parts of water and not 8 :? please explain :)

Now another question, when you get this 3/1

you say " So for every 3 parts of the original mix, we should add 1 part of pure water.
The original mix is 8 quarts. So we should add 8/3 quarts of pure water. "

let me break it down:

"So for every 3 parts of the original mix" doesn't "3" in numerator refer to water? :?

"we should add 1 part of pure water" doesn`t "1" in denominator refer to alcohol ? :?

thank you:-)
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 30 Jan 2018, 04:53
dave13 wrote:
VeritasPrepKarishma wrote:
sameerkamath21 wrote:
VeritasPrepKarishma

Hello Karishma. I recently read your blog posts on weighted averages and found the w1/w2 method and the scale method to be very useful, fast and accurate. So i've just started solving mixture problems using your techniques as much as possible. Will take sometime for me to master it. Could you please explain the how you can solve this sum using your method? Thanks



Right now the concentration of water in the mix is 4/8. We want to add pure water (8/8) to it and make the concentration of water in the final solution (5/8).

w1/w2 = (8/8 - 5/8)/(5/8 - 4/8) = 3/1

So for every 3 parts of the original mix, we should add 1 part of pure water.
The original mix is 8 quarts. So we should add 8/3 quarts of pure water.

Answer (D)


Hi VeritasPrepKarishma,

thanks for the link you shared in the previous post https://www.veritasprep.com/blog/2017/1 ... -averages/

that confusion regarding nnomenclature is clear :) now after reading the explanation which is followed by an example i have a question:

regarding this : w1/w2 = (8/8 - 5/8)/(5/8 - 4/8) = 3/1

where does it say that we need to add 8/8 pure water, it says 3 parts of alcohol and only 5 parts of water and not 8 :? please explain :)

Now another question, when you get this 3/1

you say " So for every 3 parts of the original mix, we should add 1 part of pure water.
The original mix is 8 quarts. So we should add 8/3 quarts of pure water. "

let me break it down:

"So for every 3 parts of the original mix" doesn't "3" in numerator refer to water? :?

"we should add 1 part of pure water" doesn`t "1" in denominator refer to alcohol ? :?

thank you:-)


Note that 8/8 is the concentration of water in water i.e. the water is fully water only and hence its concentration is 1. How many parts is the weight, not the concentration.
The concentration of water in the mix is 4/8 and final concentration of water is 5/8 so to ease our calculations, we write 1 as 8/8 (= 1).
The concentration of an ingredient (C1, C2 and Cavg) is expressed as Ingredient/Total (not Ingredient1/Ingredient2).

w1 and w2 are the weights of total solution (including both ingredients) in the two cases
w1/w2 = 3/1
So for every 3 parts of mix (water + alcohol), we put 1 part of water (pure water). When we do this, we will get water concentration of 5/8.
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 30 Jan 2018, 11:43
Can some one explain in detail the solution of the problem with allegation method more precisely?

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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 31 Jan 2018, 16:58
shrive555 wrote:
A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

A. 4/3
B. 5/3
C. 7/3
D. 8/3
E. 10/3


We can create the equation, in which w = the amount of added water:

4/(4 + w) = 3/5

20 = 3(4 + w)

20 = 12 + 3w

8 = 3w

8/3 = w

Answer: D
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 19 Apr 2019, 10:53
Abhishek009 wrote:
rohit8865 wrote:
shrive555 wrote:
A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

A. 4/3
B. 5/3
C. 7/3
D. 8/3
E. 10/3

Can anyone will give equation for alligation method???


For this particularproblem , algebric method is the best...

However for your convenience I am posting Alligation method -

Attachment:
1.PNG


Attachment:
2.PNG


Rest I leave for you to solve....

PS : USe alligation rule when percentage is given ( My personal opinion ) it is less time consuming and convenient....

Abhishek


Can anybody please solve the complete problem using alligation method..
am stuck after you get to 12.5..
what do you do next?
help will be much appreciated.
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many  [#permalink]

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New post 05 Aug 2019, 22:59
(4+x)/ (8+x) = 5/8

Solve
32+8x=40+5x
3x=8
x=8/3
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Re: A container holds 4 quarts of alcohol and 4 quarts of water. How many   [#permalink] 05 Aug 2019, 22:59

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