# 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) 1:3 | B) 2:3 |

C) 3:4 | D) 4:5 |

Explanation:

By the rule of alligation:

Cost of 1 kg rice of 1st kind Cost of 1 kg rice of 2nd kind

Required ratio = 60 : 90 = 2 : 3

A) 25 | B) 30 |

C) 45 | D) cannot be determined |

Explanation:

pool : kerosene

3 : 2(initially)

2 : 3(after replacement)

$\frac{RemainingQuantity}{InitialQuantity}=\left(1-\frac{ReplacedQuantity}{TotalQuantity}\right)$

(for petrol) $\frac{2}{3}=\left(1-\frac{10}{k}\right)$

=> K = 30

Therefore the total quantity of the mixture in the container is 30 liters.

A) 5:3 | B) 1:4 |

C) 4:1 | D) 9:1 |

Explanation:

Milk = 3/5 x 20 = 12 liters, water = 8 liters

If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.

Remaining milk = 12 - 6 = 6 liters

Remaining water = 8 - 4 = 4 liters

10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.

The ratio of milk and water in the new mixture = 16:4 = 4:1

If the process is repeated one more time and 10 liters of the mixture are removed,

then amount of milk removed = 4/5 x 10 = 8 liters.

Amount of water removed = 2 liters.

Remaining milk = (16 - 8) = 8 liters.

Remaining water = (4 -2) = 2 liters.

Now 10 lts milk is added => total milk = 18 lts

The required ratio of milk and water in the final mixture obtained

= (8 + 10):2 = 18:2 = 9:1.

A) 24.52 litres | B) 29.63 litres |

C) 28.21 litres | D) 25.14 litres |

Explanation:

Given that container has 50 litres of milk.

After replacing 8 litres of milk with water for three times, milk contained in the container is:

$\Rightarrow \left[50{\left(1-\frac{8}{50}\right)}^{3}\right]$

$\Rightarrow \left(50\times \frac{42}{50}\times \frac{42}{50}\times \frac{42}{50}\right)$ = 29.63 litres.

A) 4 | B) 5 |

C) 8 | D) None of these |

Explanation:

Wine Water

8L 32L

1 : 4

20 % 80% (original ratio)

30 % 70% (required ratio)

In ths case, the percentage of water being reduced when the mixture is being replaced with wine.

so the ratio of left quantity to the initial quantity is 7:8

Therefore , $\frac{7}{8}=\left[1-\frac{K}{40}\right]$

=> K = 5 Lit

A) 25 | B) 21 |

C) 12 | D) 24 |

A) 15 | B) 16 |

C) 25 | D) 31 |

Explanation:

Let quantity of mixture be x liters.

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, Where n = no of operations .

So, Quantity of Milk = $x{\left(1-\frac{6}{x}\right)}^{2}$

Given that, Milk : Water = 9 : 16

=> Milk : (Milk + Water) = 9 : (9+16)

=> Milk : Mixture = 9 : 25

Therefore, $\frac{x{\left(1-{\displaystyle \frac{6}{x}}\right)}^{2}}{x}=\frac{9}{25}$

=> x = 15 liters

A) 83.33 ml | B) 90.90 ml |

C) 99.09 ml | D) can't be determined |

Explanation:

Profit (%) = 9.09 % = 1/11

Since the ratio of water and milk is 1 : 11,

Therefore the ratio of water is to mixture = 1:12

Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml