Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options.
I came across two methods which I found very handy in solving some types of mixture problems. So I thought of sharing it with the gc community.
here they are
Rule of Alligation
It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.
Mean Price: The cost price of a unit quantity of the mixture is called mean price.
Rule of Alligation
If two quantities are mixed, then
\(\frac{Quantity of cheaper}{Quantity of dearer}=\frac{(C.P.) of dearer - (Mean Price)}{(Mean Price) - (C.P.)}\)
Taking a simple example
In what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?
C.P of cheaper=9.30
C.P of dearer=10.80
C.P of mean or mean price = 10.0
so putting the values in the formula
\(\frac{Qc}{Qd}=\frac{Cd-Cm}{Cm-Cc}\)
=\(\frac{10.80-10}{10-9.3}\)
=8:7 Ans
here they are
Rule of Alligation
It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price.
Mean Price: The cost price of a unit quantity of the mixture is called mean price.
Rule of Alligation
If two quantities are mixed, then
\(\frac{Quantity of cheaper}{Quantity of dearer}=\frac{(C.P.) of dearer - (Mean Price)}{(Mean Price) - (C.P.)}\)
Taking a simple example
In what ratio must rice at 9.30 /kg be mixed with rice at 10.80/kg so that the mixture be worth 10/kg?
C.P of cheaper=9.30
C.P of dearer=10.80
C.P of mean or mean price = 10.0
so putting the values in the formula
\(\frac{Qc}{Qd}=\frac{Cd-Cm}{Cm-Cc}\)
=\(\frac{10.80-10}{10-9.3}\)
=8:7 Ans
19 Aug 2009, 06:25
Kudos
Bookmarks
Suppose a container contains x units of liquid from which y units are taken out and replaced with water.
After n operations, the quantity of pure liquid = \([x(1-\frac{y}{x})^n]\)
Example: A container contains 40 liters of milk. From this container 4 litres of milk was taken and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
Amount of milk left after 3 operations = \([40(1-\frac{4}{40})^3]liters = (40 *\frac{9}{10}^3)\) = 29.16 liters
After n operations, the quantity of pure liquid = \([x(1-\frac{y}{x})^n]\)
Example: A container contains 40 liters of milk. From this container 4 litres of milk was taken and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
Amount of milk left after 3 operations = \([40(1-\frac{4}{40})^3]liters = (40 *\frac{9}{10}^3)\) = 29.16 liters
19 Aug 2009, 06:37
Kudos
Bookmarks
you also might wanna see a short method for mixture problems used by Economist
check out this problem
https://gmatclub.com/forum/how-much-of-the-mixture-is-to-be-removed-and-replaced-82423.html
Hope this post helps
Soon I'll be posting some mixture problems which I couldn't solve in one go ...
see ya ....
check out this problem
https://gmatclub.com/forum/how-much-of-the-mixture-is-to-be-removed-and-replaced-82423.html
Hope this post helps
Soon I'll be posting some mixture problems which I couldn't solve in one go ...
see ya ....
