# 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) 14 and 16 | B) 13 and 27 |

C) 17 and 27 | D) None of these |

Explanation:

Total no.of boys : no. of girls = 13:27

27 : 13

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%.

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) 1/3 | 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) 1.25 kg | B) 1 kg |

C) 1.5 kg | D) None of these |

Explanation:

Let the initial amount of honey in the jar was K, then

$512=K{\left(1-\frac{1}{5}\right)}^{4}$ $\left[\because 20\%=\frac{20}{100}=\frac{1}{5}\right]$

or

$512=K{\left(\frac{4}{5}\right)}^{4}$

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg

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