# 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) 1:9 | 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 **

A) 49 : 221 | B) 39:231 |

C) 94:181 | D) None of these |

Explanation:

Copper in 4 kg = 4/5 kg and Zinc in 4 kg = 4 x (4/5) kg

Copper in 5 kg = 5/6 kg and Zinc in 5 kg = 5 x (5/6) kg

Therefore, Copper in mixture = $\frac{4}{5}+\frac{5}{6}=\frac{49}{30}$ kg

and Zinc in the mixture = $\frac{16}{5}+\frac{25}{6}=\frac{221}{30}$kg

Therefore the required ratio = 49 : 221

A) 45L | B) 36.45L |

C) 40.5L | D) 42.5L |

Explanation:

General Formula:

Final or reduced concentration = initial concentration x ${\left(1-\frac{amountbeingreplacedineachoperation}{totalamount}\right)}^{n}$

where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time.

Therefore, $50\times {\left(1-\frac{5}{50}\right)}^{3}$

= 36.45 L

A) 6 ml | B) 11 ml |

C) 15 ml | D) 9 ml |

Explanation:

Let us assume that the lotion has 50% alcohol and 50% water.

ratio = 1:1

As the total solution is 9ml

alcohol = water = 4.5ml

Now if we want the quantity of alcohol = 30%

The quantity of water = 70%

The new ratio = 3:7

Let x ml of water be added

We get,

$\frac{\mathbf{4}\mathbf{.}\mathbf{5}}{\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{3}}{\mathbf{7}}$

=> x=6

Hence 6ml of water is added.

A) 71.02% | B) 76.92% |

C) 63.22% | D) 86.42% |

Explanation:

Let the milk he bought is 1000 ml

Let C.P of 1000 ml is Rs. 100

Here let he is mixing K ml of water

He is getting 30% profit

=> Now, the selling price is also Rs. 100 for 1000 ml

=> 100 : K%

= 100 : 30

10 : 3 is ratio of milk to water

=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%

A) 5 lit | B) 10 lit |

C) 15 lit | D) 20 lit |

Explanation:

Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 150 = 30 liters.

P liters of water added to the mixture to make water 25% of the new mixture.

Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).

(30 + P) = 25/100 x (150 + P)

120 + 4P = 150 + P => P = 10 liters.

A) 2 lit | B) 4 lit |

C) 1 lit | D) 3 lit |

Explanation:

Quantity of fruit juice in the mixture = 70 - [70 x (10/100) ]= 63 litres.

After adding water, juice would form 87.5% of the mixture.

Hence, if quantity of mixture after adding x liters of water, [(87.5) /100 ]*x = 63 => x = 72

Hence 72 - 70 = 2 litres of water must be added.