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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
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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
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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 ....
