## Core Concepts

- Mixture or alloys contains two or more ingredients of certain quantity mixed together to get a desired quantity. The quantity can be expressed as a ratio or percentage.
*For Ex: 1 liter of a mixture contains 250ml water and 750 ml milk. That means, 1/4 of mixture is water and 3/4 of mixture is milk. In other words, 25% of mixture is water and 75% of mixture is milk.* - Alligation 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. The cost price of unit quantity of such a mixture is called its Mean Price. Remember the rule that
*cost price of costlier ingredient > cost price of mixture > cost price of cheaper ingredient*.

## Important formulas and shortcuts for mixtures and alligations

**1) 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)

**Thus we get the important relation for alligation as**

**2) 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**

**3) If n different vessels of equal size are filled with the mixture of P and Q**

If n different vessels of equal size are filled with the mixture of P and Q in the ratio p1 : q1, p2 : q2, ……, pn : qn and content of all these vessels are mixed in one large vessel, then

Let x liters be the volume of each vessel,

Quantity of P in vessel 1 = p1 * x / (p1 + q1)

Quantity of P in vessel 2 = p2 * x / (p2 + q2)

Quantity of P in vessel n = pn * x / (pn + qn)... and so on

Similarly,

Quantity of Q in vessel 1 = q1 * x / (p1 + q1)

Quantity of Q in vessel 2 = q2 * x / (p2 + q2)

Quantity of Q in vessel n = qn * x / (pn + qn)... and so on

Therefore, when content of all these vessels are mixed in one large vessel, then

**Quantity of P / Quantity of Q = Sum of quantities of P in different vessels / Sum of quantities of Q in different vessels**

**4) If n different vessels of sizes x1, x2, …, xn are filled with the mixture of P and Q**

If n different vessels of sizes x1, x2, …, xn are filled with the mixture of P and Q in the ratio p1 : q1, p2 : q2, ……, pn : qn and content of all these vessels are mixed in one large vessel, then

Quantity of P in vessel 1 = p1 * x1/(p1 + q1)

Quantity of P in vessel 2 = p2 * x2/(p2 + q2)

Quantity of P in vessel n = pn * xn/(pn + qn)... and so on

Similarly,

Quantity of Q in vessel 1 = q1 * x1/(p1 + q1)

Quantity of Q in vessel 2 = q2 * x2/(p2 + q2)

Quantity of Q in vessel n = qn * xn/(pn + qn)

Therefore, when content of all these vessels are mixed in one large vessel

**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,**

## Comments

## Post a Comment