Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
30 Apr 2013, 14:45
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
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 .
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:
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:
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
This is my second post of Tips and Tricks, if you have missed the first one be sure to check it out
InequalitiesIn 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 96425 times ] |
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 96477 times ] |
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 [ 2.4 KiB | Viewed 96249 times ] |
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
02 May 2013, 18:03
Kudos
Bookmarks
Zarrolou-- thanks for showing some new approach !
08 Dec 2014, 13:18
Kudos
Bookmarks
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.
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.
) to solve easly any mixture problem: it's called 
