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

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10 Jun 2011, 00:08

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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?

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

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06 Apr 2016, 20:08

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?

4/3 5/3 7/3 8/3 10/3

Isn't 4 quarts to 4 quarts equivalent to a ratio of 1:1? Which is the same unit of 3:5.

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

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25 Apr 2016, 00:47

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???

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

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25 Apr 2016, 02:08

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 [ 1.87 KiB | Viewed 3737 times ]

Attachment:

2.PNG [ 4.14 KiB | Viewed 3739 times ]

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 _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

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

Show Tags

25 Apr 2016, 11:24

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

Abhi thanks 4 reply... But i could not get (50----->12.5------>62.5)part.Per me (50-12.5=37.5 as 37.5 comes from 3/8 ,similarly 5/8=67.5...fair....but not understood 50----->12.5------>62.5 also water is added so it must be 100 instead 50%) plz explain 4 me...may be i m missing something... thanks again..

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

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25 Apr 2016, 12:20

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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???

An alternate to the alligation method is the more direct/algebraic method:

Let x be the amount of water to be added.

New total amount of water = 4+x

Total amount of alcohol = 4

New total = 4+4+x=8+x

Final ratio required (for water) = 5/(5+3)=5/8

Thus, (4+x)/(8+x) = 5/8 --> solving for x you get x = 8/3.

Abhi thats what i m asking too.. as per diagram in your first post. m=12.5 & c=50 but from wherm where you calculated 62.5 because m-c inthat case is 12.5-50=...

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

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.

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?