Alligations and Mixture

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.

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
$\left ( \frac{ \text{ Quantity of Cheaper }}{\text{ Quantity of dearer }} \right ) = \left ( \frac{ \text{ C.P of Dearer (d) – Mean price (m)}}{ \text{ Mean price(m) – C.P. of cheaper(c) }} \right)$
$\frac{4}{5} = \frac{(N – 15)}{(15 – 12)}$
$\frac{4}{5} = \frac{N – 15}{3}$
$\frac{4}{5} = \frac{N – 15}{3}$
N = Rs. 17.4 per Kg

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

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

Example :

Question 1: 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