mixtures and alligations problems

Alligation rule is used to determine the exact proportion to mix two ingredients of any price to obtain a mixture with estimated cost.

Mean price: Cost price of the mixture per unit.

suppose a container contains x units of liquid out of which y units are replaced then quantity of liquid after n such operations left in container would be x(1-y/x)^n 

Alligation rule:

Quantity of cheap ingredient(b) : Quantity of costly ingredient (a)= x2-x1 : x1-x0

point to remember:
cost price of mixture always lies between two costs.

Consider the graph Quantity v/s Cost , with this method one can solve problems with more ease

A: Type of cheaper ingredient.

B: Type of costly ingredient.
x0: cost of cheaper ingredient.
x1: cost of mixture.
x2: cost of costly ingredient.
Quantity of cheap ingredient : Quantity of costly ingredient = x2-x1 : x1-x0


Mixtures and alligations solved problems:
Question:
A tank of capacity 100 litres is filled with milk and 1/4 portion of tank (i.e 25 litres) is replaced with water for 4 times, what would be the quantity of milk left?
solution:
=100(1-25/100)^4
=100(3/4)^4
=100(81/256)
=31.6 litres

Question 2:
A tank is filled with a solution, 1/4 portion of tank is replaced with water for 3 times, what would be the solution left?
solution:
=1(1-1/4)^3
=(3/4)^3
=27/64

Question 3:
cost of ingredient A is 100, cost of ingredient B is 200, in what ratio the ingredients to be mixed to obtain a mixtiure with cost of 170??
solution
Quantity of cheap ingredient(b) : Quantity of costly ingredient (a)= x2-x1 : x1-x0

b:a=200-170:170-100

b:a=30:70

b:a=3:7

so 3 units of A and 7 units of B gives a mixture of cost 170

Question 4:

Cost of an ingredient (I1) is 160, cost of mixture is 100, find cost of another ingredient (I2) when ratio of quantities in mixture I1:I2 is 4:6

solution:

As cost of mixture always lies between two ingredients, 

cost of another ingredient here would be less than 100.

Quantity of cheap ingredient(b)(I2) : Quantity of costly ingredient (a)(I1)= x2-x1 : x1-x0

b:a=160-100:100-x0

6:4=60:100-x0

100-x0=40
x0=60
cost of cheaper ingredient is 60

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