1. Rule of Alligation:

If two ingredients are mixed, then

(Quantity of cheaper/Quantity of dearer) = (C.P. of dearer - Mean Price) / (Mean price - C.P. of cheaper)

We present as under:

(Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c)

2.Suppose a container contains x of liquid from which y units are taken out and replaced by water.

After n operations, the quantity of pure liquid = [x(1-(y/x))^n] units.

### General Tips

1. 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 desired price.

Mean Price:

The cost of a unit quantity of the mixture is called the mean price.

3. Rule Of Alligation

Given

Quantity of cheaper ingredient = qc

Cost price of cheaper ingredient = pc

Quantity of dearer or costlier ingredient = qd

Cost price of costlier or dearer ingredient = pd

Consider, mean price of mixture as pm and quantity of mixture as qm.

We know, qm = qc + qd

Then we get,

(qc * pc + qd * pd) = qm * pm = (qc + qd) * pm

=> qc ( pm – pc) = qd (pd – pc)

=> qc / qd = (pd – pc) / ( pm – pc)

4. Quantity of ingredient to be added to increase the content of ingredient in the mixture to y%

If P liters of a mixture contains x% ingredient in it. Find the quantity of ingredient to be added to increase the content of ingredient in the mixture to y%.

Let the quantity of ingredient to be added = Q liters

Quantity of ingredient in the given mixture = x% of P = x/100 * P

Percentage of ingredient in the final mixture = Quantity of ingredient in final mixture / Total quantity of final mixture.

Quantity of ingredient in final mixture = [x/100 * P] + Q = [ P*x + 100 * Q] / 100

Total quantity of final mixture = P + Q

=> y/100 = [[ P*x + 100 * Q] / 100]/[P + Q]

=> y[P + Q] = [P*x + 100 * Q]

The quantity of ingredient to be added Q = P(y-x)/(100-y)

5) Quantity of ingredient to be added to change the ratio of ingredients in a mixture

In a mixture of x liters, the ratio of milk and water is a : b. If the this ratio is to be c : d, then the quantity of water to be further added is:

In original mixture

Quantity of milk = x * a/(a + b) liters

Quantity of water = x * b/(a + b) liters

Let quantity of water to be added further be w litres.

Therefor in new mixture:

Quantity of milk = x * a/(a + b) liters → Equation(1)

Quantity of water = [x * b/(a + b) ] + w liters → Equation (2)

=> c / d = Equation (1) / Equation (2)

Quantity of water to be added further,

w = [x *(ad-bc)/c(a+b)]