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Arithmetic II

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Mixtures & Alligations

Mixtures Alligations

MODULES

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Weighted Average & Alligations
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Examples 3 to 6
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Examples 7 to 13
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Successive Replacement & Examples 14 to 17
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Hybrid, Evaporation & Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Averages & Alligations 1
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Averages & Alligations 2
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Averages & Alligations 3
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Averages & Alligations 4
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PRACTICE

Mixtures & Alligations : Level 1
Mixtures & Alligations : Level 2
Mixtures & Alligations : Level 3
ALL MODULES

CAT 2025 Lesson : Mixtures & Alligations - Concepts & Cheatsheet

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Note: The video for this module contains a summary of all the concepts covered in the Mixtures & Alligations lesson. The video would serve as a good revision. Please watch this video in intervals of a few weeks so that you do not forget the concepts. Below is a cheatsheet that includes all the formulae but not necessarily the concepts covered in the video.

8. Cheatsheet

1)1)1) x4=x1w1+x2w2+...+xnwnw1+w2+...+wnx_{4} = \dfrac{x_{1}w_{1} + x_{2}w_{2} + ... + x_{n}w_{n}}{w_{1} + w_{2} + ... + w_{n}}x4​=w1​+w2​+...+wn​x1​w1​+x2​w2​+...+xn​wn​​.

2)2)2) w1w2=(x2−xA)(xA−x2)\dfrac{w_{1}}{w_{2}} = \dfrac{(x_{2} - x_{A})}{(x_{A} - x_{2})}w2​w1​​=(xA​−x2​)(x2​−xA​)​



333) Where I\bm{I}I is the initial quantity of non-replaced liquid,
r%\bm{r\%}r% is the replacement percentage,
n\bm{n}n is the number of replacements,
F\bm{F}F is the final quantity of the non-replaced liquid,

F=I×(1−r100)nF = I \times \left(1 - \dfrac{r}{100} \right)^{n}F=I×(1−100r​)n

444) If the quantity replaced is provided (for example, in litres) instead of percentage,

F=I×(1−Replacement QuantityTotal Quantity)nF = I \times \left(1 - \dfrac{\text{Replacement Quantity}}{\text{Total Quantity}} \right)^{n}F=I×(1−Total QuantityReplacement Quantity​)n

555) Where the quantities of 222 mixtures of different unknown concentrations are a\bm{a}a and b\bm{b}b , and a quantity of x\bm{x}x is removed from each of the mixtures and placed in the other mixtures, and this results in both the mixtures having the same concentrations, then

x=aba+bx = \dfrac{ab}{a + b}x=a+bab​
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