# 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) Rs. 169.50 | B) Rs.1700 |

C) Rs. 175.50 | D) Rs. 180 |

Explanation:

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

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

x-153/22.50 = 1 => x - 153 = 22.50 => x=175.50.

Hence, price of the third variety = Rs.175.50 per kg.

A) 18 litres | B) 24 litres |

C) 32 litres | D) 42 litres |

Explanation:

Let the quantity of the wine in the cask originally be x litres

Then, quantity of wine left in cask after 4 operations =$\left[x{\left(1-\frac{8}{x}\right)}^{4}\right]$litres

$\therefore \left[\frac{x{\left(1-{\displaystyle \frac{8}{x}}\right)}^{4}}{x}\right]=\frac{16}{81}$

$\Rightarrow {\left[1-\frac{8}{x}\right]}^{4}={\left(\frac{2}{3}\right)}^{4}$

$\Rightarrow x=24$

A) 10 | B) 20 |

C) 21 | D) 25 |

Explanation:

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively

Quantity of A in mixture left =$\left(7x-\frac{7}{12}\times 9\right)=\left(7x-\frac{21}{4}\right)$ litres.

Quantity of B in mixture left = $\left(5x-\frac{5}{12}\times 9\right)=\left(5x-\frac{15}{4}\right)$litres.

$\frac{\left(7x-{\displaystyle \frac{21}{4}}\right)}{\left(5x-{\displaystyle \frac{15}{4}}\right)+9}=\frac{7}{9}$

$\Rightarrow \frac{28x-21}{20x+21}=\frac{7}{9}$

$\Rightarrow x=3$

So, the can contained 21 litres of A.

A) 400 kg | B) 560 kg |

C) 600 kg | D) 640 kg |

Explanation:

By the rule of alligation:

Profit of first part Profit of second part

So, ratio of 1st and 2nd parts = 4 : 6 = 2 : 3.

Therefore, Quantity of 2nd kind = (3/5 x 1000)kg = 600 kg.

A) 1/3 | B) 1/4 |

C) 1/5 | D) 1/7 |

Explanation:

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = $\left(3-\frac{3x}{8}+x\right)$litres.

Quantity of syrup in new mixture = $\left(5-\frac{5x}{8}\right)$ litres.

$\left(3-\frac{3x}{8}+x\right)=\left(5-\frac{5x}{8}\right)$

=> 5x + 24 = 40 - 5x

=> 10x = 16 => x = 8/5

So, part of the mixture replaced = $\left(\frac{8}{5}\times \frac{1}{8}\right)$ = 1/5.

A) 10 liters | B) 20 liters |

C) 30 liters | D) 40 liters |

Explanation:

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters

Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)

Thus, (30 + P) = 25% of (150 + P)

Solving, we get P = 10 liters

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) 25 | B) 21 |

C) 12 | D) 24 |