# 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) 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) 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

A) 75 liters | B) 100 liters |

C) 150 liters | D) 120 liters |

Explanation:

$\frac{Wine\left(left\right)}{Water\left(added\right)}=\frac{343}{169}$

It means $\frac{Wine\left(left\right)}{Wine(initialamount)}=\frac{343}{512}$ $\left[\because 343+169=512\right]$

Thus ,

$343x=512x{\left(1-\frac{15}{K}\right)}^{3}$

$\Rightarrow \frac{343}{512}={\left(\frac{7}{8}\right)}^{3}={\left(1-\frac{15}{k}\right)}^{3}$

=> K = 120

Thus the initial amount of wine was 120 liters.

A) 7:3 | B) 5:3 |

C) 8:5 | D) 2:7 |

Explanation:

$W1:A1W2:A2.....WN:AN$

2 : 3 4 : 5 5 : 7

$\frac{W1}{W1+A1}=\frac{2}{5}\frac{W2}{W2+A2}=\frac{4}{9}\frac{WN}{WN+AN}=\frac{5}{12}$

= 72/180

= 80/180

= 75/180

=> 5 : 3

Therefore, the ratio is **5: 3**

A) 1 litre | B) 3 litres |

C) 4 litres | D) 5 litres |

Explanation:

Total quantity of mixture = 75 litre

Milk : Water = 4 : 1

3 : 1

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) 80% | B) 70% |

C) 75% | D) 62% |

Explanation:

Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm.

After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and weight of a mixture = 60 + 15 = 75 gm.

So % of water = 100 x 60/75 = 80%.