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# Two important formulas for mixture problems

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Intern
Joined: 11 Aug 2009
Posts: 10
Two important formulas for mixture problems  [#permalink]

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Updated on: 19 Aug 2009, 06:52
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9
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

Originally posted by joebloggs on 19 Aug 2009, 06:10.
Last edited by joebloggs on 19 Aug 2009, 06:52, edited 2 times in total.
Intern
Joined: 11 Aug 2009
Posts: 10

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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
##### General Discussion
Intern
Joined: 11 Aug 2009
Posts: 10
Re: Two important formulas for mixture problems  [#permalink]

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19 Aug 2009, 06:37
3
you also might wanna see a short method for mixture problems used by Economist
check out this problem

http://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 ....
Senior Manager
Joined: 18 Jun 2009
Posts: 310
Location: San Francisco
Re: Two important formulas for mixture problems  [#permalink]

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19 Aug 2009, 12:40
nice post .. thanks .. +1
Manager
Joined: 16 Apr 2009
Posts: 152
Schools: Ross
Re: Two important formulas for mixture problems  [#permalink]

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19 Aug 2009, 12:53
Keep up the good work !
Intern
Joined: 14 Aug 2019
Posts: 3
Re: Two important formulas for mixture problems  [#permalink]

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14 Aug 2019, 02:59
Manager
Joined: 01 Jan 2017
Posts: 57
WE: General Management (Consulting)
Two important formulas for mixture problems  [#permalink]

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28 Aug 2019, 00:40
Quote:
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

For those who love a more visual approach, consider the following, which may be even simpler.

Just subtract cheaper and dearer from the mean and find the difference.

---9.3----10.8--
-----\----/------
-------10-------
------/---\------
---0.8---0.7---
Two important formulas for mixture problems   [#permalink] 28 Aug 2019, 00:40
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