Mixture with Alligation

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Mixture with Alligation

by nadib002 » Thu Apr 28, 2011 9:06 pm
A vessel contains 5 parts of milk and 1 part water. how much of the mixture is to be removed and replaced by water to make the resulting mixture 1/2 water and 1/2 milk ?

The answer is 2 2/5.

I used the alligation method wherein I took the ratio of water and solved for it. I got 2/5 ( the amount of water that needs to be removed). But i do-not seem to understand why we add 2 for the milk.

There is a post on the forum on the same question but i would like to know a way to solve using "Alligation" method

Could you please help

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by Stuart@KaplanGMAT » Fri Apr 29, 2011 12:10 am
Hi!

I've never heard of the "alligation" method - does it involve feeding part of the mixture to alligators? Sounds dangerous!

As an aside, this question is seriously flawed and impossible to answer in its current form, since you can't answer a "how much" question unless you have an actual quantity. If all we know is the ratio of water to milk, then all we can answer is "what percent of the mixture" must be replaced by water, not "how much of the mixture".
nadib002 wrote:A vessel contains 5 parts of milk and 1 part water. how much of the mixture is to be removed and replaced by water to make the resulting mixture 1/2 water and 1/2 milk ?

The answer is 2 2/5.

I used the alligation method wherein I took the ratio of water and solved for it. I got 2/5 ( the amount of water that needs to be removed). But i do-not seem to understand why we add 2 for the milk.

There is a post on the forum on the same question but i would like to know a way to solve using "Alligation" method

Could you please help
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by nadib002 » Fri Apr 29, 2011 6:16 am
@Stuart

Thank you for the post.

Here's the alligation method.

When combining a lower percentage (L) with a higher percentage (H) to achieve a goal percentage (G):

The proportion needed of L = H-G (the positive difference between the higher percentage and the goal percentage)
The proportion needed of H = G-L (the positive difference between the lower percentage and the goal percentage)

[Courtesy GMATGuruNY]

I used the above method to solve the problem wherein I took

Initial percentage of water = (5/6)*100

Goal percentage of water = (1/2)%100

Amount removed = 100% water

Am I missing something here?

Here's the link to the original post:
https://www.beatthegmat.com/please-can-s ... 12244.html


Please help

Thank you

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by GMATGuruNY » Fri Apr 29, 2011 6:42 am
nadib002 wrote:A vessel contains 5 parts of milk and 1 part water. how much of the mixture is to be removed and replaced by water to make the resulting mixture 1/2 water and 1/2 milk ?

The answer is 2 2/5.

I used the alligation method wherein I took the ratio of water and solved for it. I got 2/5 ( the amount of water that needs to be removed). But i do-not seem to understand why we add 2 for the milk.

There is a post on the forum on the same question but i would like to know a way to solve using "Alligation" method

Could you please help
As Stuart noted, the problem should give a volume.
Let's assume that the total volume is 5 liters milk + 1 liter water = 6 liters.

Using alligation:

Fraction of water in the original solution = 1/6
Fraction of water in the water to be added = 1
Desired fraction of water in the final mixture = 1/2

Proportion needed of the original solution = 1 - 1/2 = 1/2.
Proportion needed of the water to be added = 1/2 - 1/6 = 2/6.

Required ratio of solution to water = (1/2) : (2/6) = (3/6) : (2/6) = 3:2.

The sum of the elements in the ratio is 3+2 = 5.
The part of the ratio attributed to the added water = 2.
Thus, the added water = part/whole = 2/5 of the 6 liters.
(2/5) * 6 = 12/5.
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by nadib002 » Fri May 06, 2011 7:08 am
@Mitch

Thank you