# Alligation or Mixture Questions

**FACTS AND FORMULAE FOR ALLIGATION OR MIXTURE QUESTIONS**

**I. 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 a desired price.

**II. Mean Price :** The cost price of a unit quantity of the mixture is called the mean price.

**III. Rule of Alligation :** Suppose Rs.x per unit be the price of first ingradient mixed with another ingradient (cheaper) of price Rs.y per unit to form a mixture whose mean price is Rs. z per unit, then

Quantity of cheaper : Quantity of dearer

= ( C.P of dearer - Mean Price ) : ( Mean Price - C. P of cheaper )

= ( x- z ) : ( z - y )

**IV.** Suppose a container contains x units of liquid from which y units are taken out and replaced by water. After operations , the quantity of pure liquid = $x{\left(1-\frac{y}{x}\right)}^{n}$ units.

A) 11 lit | B) 22 lit |

C) 33 lit | D) 44 lit |

Explanation:

Let the capacity of the pot be 'P' litres.

Quantity of milk in the mixture before adding milk = 4/9 (P - 8)

After adding milk, quantity of milk in the mixture = 6/11 P.

6P/11 - 8 = 4/9(P - 8)

10P = 792 - 352 => P = 44.

The capacity of the pot is 44 liters.

A) 65 liters | B) 91 liters |

C) 38 liters | D) None of these |

Explanation:

Milk Water

74% 26% (initially)

76% 24% ( after replacement)

Left amount = Initial amount $\left(1-\frac{replacedamount}{totalamount}\right)$

24 = 26$\left(1-\frac{7}{k}\right)$

=> k = 91

A) 193 : 122 | B) 97 : 102 |

C) 115 : 201 | D) 147 : 185 |

Explanation:

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

i.e (35,70), (42,63), (45,60)

So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

But in the question asked the ratio of water to milk = 193 : 122

A) 26.34 litres | B) 27.36 litres |

C) 28 litres | D) 29.16 litres |

Explanation:

Amount of milk left after 3 operations

$\left(40{\left(1-\frac{4}{40}\right)}^{3}\right)=\left(40*\frac{9}{10}*\frac{9}{10}*\frac{9}{10}\right)=29.16$ litres

A) 81.92L | B) 96L |

C) 118.08L | D) None of these |

Explanation:

The amount of petrol left after 4 operations = $200\times {\left(1-\frac{40}{200}\right)}^{4}=$81.92

Hence the amount of kerosene = 200 - 81.92 = 118.08 liters

A) 1:9 | B) 9:1 |

C) 3:7 | D) 7:3 |

Explanation:

Ratio of Milk and water in a vessel A is 4 : 1

Ratio of Milk and water in a vessel B is 3 : 2

Ratio of only milk in vessel A = 4 : 5

Ratio of only milk in vessel B = 3 : 5

Let 'x' be the quantity of milk in vessel C

Now as equal quantities are taken out from both vessels A & B

=> 4/5 : 3/5

x

3/5-x x - 4/5

=> $\frac{{\displaystyle \frac{\mathbf{3}}{\mathbf{5}}\mathbf{-}\mathbf{x}}}{\mathbf{x}\mathbf{-}{\displaystyle \frac{\mathbf{4}}{\mathbf{5}}}}$ **= $\frac{\mathbf{1}}{\mathbf{1}}$** (equal quantities)

=> x = 7/10

Therefore, quantity of milk in vessel C = 7

=> Water quantity = 10 - 7 = 3

Hence the ratio of milk & water in vessel 3 is **7 : 3 **