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Tips and Tricks: Mixtures

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Tips and Tricks: Mixtures [#permalink] New post 30 Apr 2013, 13:45
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Hi guys!

This is my second post of Tips and Tricks, if you have missed the first one be sure to check it out :arrow: Inequalities

In this one I will show you a method (disclamer: I did not invente it :roll: ) to solve easly any mixture problem: it's called Alligation.

It uses a simple table to solve any mixture problem, every answer to such problems can be obtain by looking at this table .

Attachment:
alli.png
alli.png [ 6.15 KiB | Viewed 4067 times ]

Please note: the X concentration is the highest, the Y is the lowest

The results that you get by subtracting, as I show you in the table, are the ratios of the substances in the desired mixture.
RATIO\frac{X}{Y}=\frac{Desired-Y}{X-Desired}

An exmple will explain better than any of my words because this method is really simple to use. I took the following questions from here if you want to get some practice you can try some of those.

1)Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue.
If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of this mixture is X ?
(A) 10%
(B) 33.33 %
(C) 40%
(D) 50%
(E) 66.66 %

The question asks for the ryegrass so your table should look like this:
Attachment:
all1.png
all1.png [ 2.28 KiB | Viewed 4064 times ]

Solution: The final raio is\frac{X}{Y}=\frac{5}{10} (or \frac{1}{2}) so for every 1 part of X 2 parts of Y will be in the final mixture
So for a 3 kg mixture (for example)=> 1X and 2Y => X=33% of the total B

This table can be used in other ways also, and this question is an example:
2)How many liters of pure alcohol must be added to a 100-liter solution that is 20 percent alcohol in order to produce a solution that is 25 percent alcohol?
(A) 7/2
(B) 5
(C) 20/3
(D) 8
(E) 39/4

Your table:
Attachment:
all2.png
all2.png [ 2.4 KiB | Viewed 4064 times ]

Final ratio: \frac{X}{Y} = \frac{5}{75}
We know that Y is 100 liter so \frac{X}{100}=\frac{5}{75} X=\frac{20}{3} C
Easy!

As you see mixture problems start to look very easy if you consider this method, and for sure all this will save you valuable time 8-)
Hope you guys like it

Cheers
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Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
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Re: Tips and Tricks: Mixtures [#permalink] New post 02 May 2013, 17:03
Zarrolou-- thanks for showing some new approach !
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Re: Tips and Tricks: Mixtures [#permalink] New post 29 Jun 2014, 06:32
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Tips and Tricks: Mixtures   [#permalink] 29 Jun 2014, 06:32
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