# Alligation or Mixture Aptitude Questions, Problems, Tricks, Formulas and Shortcuts

## Alligation or Mixture Formulas

A: It is a rule to find respective prices should be mixed to give a mixture at a given price.

B: The means or average price of a mixture when the price of two or more ingredients which may be mixed together and the proportion in which they are mixed are given,

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

Or in easy term Alligation is a rule which is used to solve the problems related to mixture and its ingredient.

Alligation Rule: Suppose, Rs. D per unit be the price of first ingredient (superior quality) mixed with another ingredient (cheaper quality) of price Rs.C per unit to form a mixture whose mean price is Rs.M per unit, then the two ingredients must be mixed in the ratio:

$\frac{Quantity \, of\, Cheaper}{Quantity\, of\, Superior}$ $\frac{CP\,Superior - Mean\, Price}{Mean\, Price - CP\, of\, Cheaper}$

i.e. the two ingredients are to be mixed in the inverse ration of the two different are to be mixed and the mean price. The above rule may be represented schematically as under: Explanation: Suppose x kg of cheaper quality is mixed with y kg of superior quality.

Price of the cheaper ingredient – Rs.cx Price of superior ingredient = Rs.dy

$\therefore$ Price of mixture = Rs.(cx + dy)

And quantity of mixture = (x + Y) kg.

$\therefore$ Price of mixture/kg = Rs. (cx + dy)

and quantity of mixture = (x + y) kg.

$\therefore$ Price of mixture/kg = Rs.$(\frac{cx \,+\,dy}{x\, +\, y})$

$\therefore \,(\frac{cx \,+\,dy}{x\, +\, y})$ = m

$\Rightarrow$ cx = dy = mx = my

$\Rightarrow$ dy - my = mx - cx

$\Rightarrow$ y(d - m) = x(m - c)

$\Rightarrow \frac{x}{y} = \frac{d - m}{m - c}$

## Important Examples and Questions of Alligation or Mixture

Example 1: In what ration two varieties of tea one costing Rs.27 per kg and the other costing Rs.32 per kg should be blended to produce a blended variety of tea worth Rs.30 per kg. how much should be the quantity of second variety of tea, if the first variety is 60 kg.

Solution: The required ration of the two varieties of tea is 2 : 3, i.e.

$\frac{Quantity\, of\, Cheaper\, Tea}{Quantity\, of\, Superior\, Tea} = \frac{2}{3}$

$\therefore$ Quantity of superior tea = $\frac{60\times 3}{2}\,=\,90kg$

Example 2: Sugar at Rs.15 per kg is mixed with suger at Rs.20 per kg in the ration 2 : 3. Find the price per kg of the mixture.

Solution: Let the mean of the mixture be Rs.x. $\frac{Quantity\, of \, Cheaper\, Sugar}{Quantity\, of \, Dearer\, Sugar} = \frac{20-x}{x-15}$

$\therefore \frac{20-x}{x-15} = \frac{2}{3} \Rightarrow 60-3x = 2x-30$

$\Rightarrow 5x = 90\,or\, x = 18$

Thus the price per kg of the mixture is Rs. 18

## Previous Year Questions on Alligation or Mixture

• Al ligation or Mixture aptitude

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