# Tips And Tricks And Shortcuts For Alligation And Mixtures

## Tips And Tricks For Alligation And Mixtures

Tips and Tricks for Alligation and Mixtures will be discussed on this page for solution of problems regarding the mixing of ingredients in a shortcut way. When, two or more ingredients are mixed together to get a desired quantity, this quantity can be expressed as ratio or percentage.

## Type 1. Tips and Tricks for Alligation and Mixtures

In most of these questions , you are supposed to find the ratio , or any one of the values either M(mean price) , or C(C.P of cheaper) or D(C.P of dearer) where the other two values and the ratio is given.

To find that , we have the best trick for you.

When two commodities are mixed then,

$\left ( \frac{ \text{ Quantity of Cheaper }}{\text{ Quantity of dearer }} \right ) = \left ( \frac{ \text{ C.P of Dearer (d) – Mean price (m)}}{ \text{ Mean price(m) – C.P. of cheaper(c) }} \right)$

This equation can also be developed using the pictorial diagram.

### Question 1:

Two varieties of wheat are mixed in the ratio of 4:5. The price of the mixture is ₹15 per Kg. The price of the variety having lesser weight is ₹12 per Kg. Calculate the price of the other variety.

### Solution:

First of all we will identify and substitute the values in the diagram

Now substituting this in the formula

$\left ( \frac{ \text{ Quantity of Cheaper }}{\text{ Quantity of dearer }} \right ) = \left ( \frac{ \text{ C.P of Dearer (d) – Mean price (m)}}{ \text{ Mean price(m) – C.P. of cheaper(c) }} \right)$

$\frac{4}{5} = \frac{(N – 15)}{(15 – 12)}$
$\frac{4}{5} = \frac{N – 15}{3}$

So , 5N – 75 = 12

N = Rs. 17.4 per Kg

## Type 2: Shortcuts, Tips and Tricks for Alligation and Mixtures

Calculate quantity of pure Liquid after ‘n’ successive operations:

If a Container contains ‘x’ units of pure liquid , and we replace the liquid with ‘y’  units of water :

Then after ‘n’ successive operations, the units of pure liquid left is

$\left ( x \left( 1 – \frac{y}{x} \right)n \right)$

Below  is an example to explain this concept.

### Question:

A vessel contains 60L of milk, out of which 15L litres of milk is taken out and replaced by water. This process is repeated two times. Find the amount of milk left in the container.

### Solution:

From the question we have ,

Total Milk (x)  = 60L

Milk taken out in one round (y) = 15L

No. of rounds (N) = 2

So , Using the above formula ,

Amount of milk left in the container

=$\left ( x \left( 1 – \frac{y}{x} \right)n \right)$

=$\left ( 60 \left( 1 – \frac{15}{60} \right)2 \right)$

=$\left ( 60 \left( \frac{3}{4} \right)2 \right)$

Solving this we get the answer as 33.75L