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# Tips and Tricks: Mixtures

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Status: Far, far away!
Joined: 02 Sep 2012
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Location: Italy
Concentration: Finance, Entrepreneurship
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Tips and Tricks: Mixtures  [#permalink]

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30 Apr 2013, 13:45
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102
Hi guys!

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

In this one I will show you a method (disclamer: I did not invente it ) 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 [ 6.15 KiB | Viewed 49058 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 [ 2.28 KiB | Viewed 49339 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

 Attachment: all2.png [ 2.4 KiB | Viewed 49152 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
Hope you guys like it

Cheers
_________________

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Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
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Senior Manager
Joined: 10 Apr 2012
Posts: 268
Location: United States
Concentration: Technology, Other
GPA: 2.44
WE: Project Management (Telecommunications)
Re: Tips and Tricks: Mixtures  [#permalink]

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02 May 2013, 17:03
Zarrolou-- thanks for showing some new approach !
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Joined: 17 Jul 2014
Posts: 1
Re: Tips and Tricks: Mixtures  [#permalink]

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08 Dec 2014, 12:18
1
Zarrolou- Thanks for the approachThree grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?

A. y + 3z
B. (y +z) / 4
C. 2y + 3z
D. 3y + z
E. 3y + 4.5z

Can you please let me know how I could solve this question with the help of your table.
Basically I need the table for more than 2 unknown mixtures.
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Re: Tips and Tricks: Mixtures  [#permalink]

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08 Dec 2014, 12:20
sagarswamy wrote:
Zarrolou- Thanks for the approachThree grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?

A. y + 3z
B. (y +z) / 4
C. 2y + 3z
D. 3y + z
E. 3y + 4.5z

Can you please let me know how I could solve this question with the help of your table.
Basically I need the table for more than 2 unknown mixtures.

You can check various methods to solve this question here: three-grades-of-milk-are-1-percent-2-percent-and-3-percent-fat-126122.html

Hope it helps.
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Joined: 16 Apr 2016
Posts: 20
Re: Tips and Tricks: Mixtures  [#permalink]

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03 May 2016, 00:46
Wow, this makes mixture problems so much easier. That's one of the sections i've been having problems understanding the logic behind them.
First i think i need to look at these graphs to get used to using this method, but after a while i think it will just pop from the back of my head.
THAnK YOU!
Intern
Joined: 07 Jan 2018
Posts: 11
Re: Tips and Tricks: Mixtures  [#permalink]

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28 Sep 2018, 05:46
In first example, how did you get the answer as 33℅ after getting x/y=1/2?

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 52294
Tips and Tricks: Mixtures  [#permalink]

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28 Sep 2018, 08:04
mittu wrote:
In first example, how did you get the answer as 33℅ after getting x/y=1/2?

Posted from my mobile device

x/y = 1/2;

x/(x + y) = 1/(1 + 2) = 1/3 = 0.333....

You can find detailed discussion of that question here: https://gmatclub.com/forum/seed-mixture ... 60580.html
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Tips and Tricks: Mixtures &nbs [#permalink] 28 Sep 2018, 08:04
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# Tips and Tricks: Mixtures

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