###### Formulae

1. 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 desired price.

2. Mean Price:

The cost of a unit quantity of the mixture is called the mean price.

3. Rule of Alligation:

If two ingredients are mixed, then

\( {{Quantity of cheaper} \over {Quantity of dearer}} \) = \( {{C.P. of dearer - Mean Price} \over {Mean price - C.P. of cheaper}} \)

We present as under:

∴(Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c).

4. Suppose a container contains x of liquid from which y units are taken out and replaced by water.

After n operations, the quantity of pure liquid = \( [x {({1 - {y \over x}})^n}] \) units.

###### Exercise : Alligation or Mixture MCQ Questions and Answers

**1. **
How much coffee of variety A, costing Rs. 5 a kg should be added to 20 kg of Type B coffee at Rs. 12 a kg so that the cost of the two coffee variety mixture be worth Rs. 7 a kg?

A. |

B. |

C. |

D. |

**Correct Answer : D. 50 kg **

**Description :**

As per the rule of alligation,

Quantity of Dearer: Quantity of Cheaper = (12-7) : (7-5) = 5:2

Quantity of Variety A coffee that needs to be mixed ⇒ 5:2 = x:20

⇒ x =50 kg

**2. **
The ratio of milk and water in three samples is 1 ∶ 3, 3 ∶ 5 and 11 ∶ 5. A mixture containing equal quantities of all three samples is made. What will be the ratio of milk and water in the new mixture?

A. |

B. |

C. |

D. |

**Correct Answer : A. 7 : 9 **

**Description :**

**Method 1 -**

Let ‘x’ quantity of each of three mixtures to be mixed

Quantity of milk in the new mixture = (1/4)x + (3/8)x + (11/16)x = 21x/16

Quantity of water in the new mixture = (3/4)x + (5/8)x + (5/16)x = 27x/16

∴ Required ratio = (21x/16) ∶ (27x/16) = 7 ∶ 9

**Method 2 -**

1 + 3 = 4, 3 + 5 = 8, 11 + 5 = 16

Let the quantity of each sample LCM(4, 8, 16) = 16

Milk in the first sample

⇒ 16 × 1/(1 + 3)

⇒ 4

Water in the first sample = 16 - 4 = 12

Milk in the second sample

⇒ 16 × 3/8

⇒ 6

Water in the second sample = 16 - 6 = 10

Milk in the third sample

⇒ 16 × 11/16

⇒ 11

Water in the second sample = 16 - 11 = 5

Ratio of milk and water in the new mixture

⇒ ( 4 + 6 + 11)/(12 + 10 + 5)

⇒ 21/27 = 7/9

∴ Required ratio is 7 : 9.

**3. **
A vessel full of orange juice contains 40% orange pulp. A part of juice is replaced by another juice containing 19% orange pulp and now the percentage of orange pulp is found to be 26%. What quantity of juice is replaced?

A. |

B. |

C. |

D. |

**Correct Answer : A. 2:3 **

**Description :**

Concentration of orange pulp in 1st vessel = 40% Concentration of orange pulp in 2nd vessel = 19% After the mixing, Concentration of orange pulp in the mixture = 26% By rule of alligation,

Concentration of orange pulp in 1st Concentration of orange pulp in 2nd Vessel Vessel 40% 19% 26% 26-19=7 40-26=14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 i.e., 2/(1+2)=2/3 part of the juice is replaced.

**4. **
Cost of two types of pulses is Rs.15 and Rs, 20 per kg, respectively. If both the pulses are mixed together in the ratio 2:3, then what should be the price of mixed variety of pulses per kg?

A. |

B. |

C. |

D. |

**Correct Answer : C. Rs. 18 per kg **

**Description :**

Let the cost of mixed variety of pulse be Rs. x

As per the alligation rule,

2:3 = (20-x) : (x-15)

⇒ 2x+3x = 60+30

⇒ 5x = 90

⇒ x = 18

**5. **
A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

A. |

B. |

C. |

D. |

**Correct Answer : D. 10 **

**Description :**

Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 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 * (150 + P)

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

**6. **
Two solutions S1 and S2 contain whisky and soda in the ratio 2 : 5 and 6 : 7 respectively. In what ratio these solutions be mixed to get a new solution S3, containing whisky and soda in the ratio 5 : 8 ?

A. |

B. |

C. |

D. |

**Correct Answer : A. 7:9 **

**Description :**

Let the amount taken from S1 be 7x

And amount taken from S2 be 13y

(2x + 6y)/(5x + 7y) = 5/8

16x + 48y = 25x + 35y

9x = 13y

x/y = 13/9

Actual ratios of amounts

= 7x/13y

= (7/13) * (13/9)

= 7/9

**7. **
A mixture is made by mixing alcohol and water in the ratio 9 ∶ 7. If ‘x’ litres of alcohol and ‘3x’ litres of water is mixed in 80 litres of mixture, the new ratio becomes 13 ∶ 14. Find the quantity of new mixture.

A. |

B. |

C. |

D. |

**Correct Answer : A. 108 litres **

**Description :**

Sum of first ratio = 9 + 7 = 16
Quantity of alcohol in 80 litres of mixture = 9/16 × 80 = 45 litres

Quantity of water in 80 litres of mixture = 7/16 × 80 = 35 litres

When ‘x’ litres of alcohol and ‘3x’ litres of water is added,

Quantity of alcohol in new mixture = 45 + x

Quantity of water in new mixture = 35 + 3x

But, ratio of new mixture = 13 ∶ 14

⇒ (45 + x)/(35 + 3x) = 13/14

⇒ 14(45 + x) = 13(35 + 3x)

⇒ 630 + 14x = 455 + 39x

⇒ 39x – 14x = 630 – 455

⇒ x = 175/25 = 7

∴ Total quantity of new mixture = 45 + x + 35 + 3x = 80 + 4(7) = 80 + 28 = 108 litres

**8. **
A milkman claims to sell milk at its cost price, still, he is making a profit of 30% since he has mixed some amount of water in the milk. What is the % of milk in the mixture?

A. |

B. |

C. |

D. |

**Correct Answer : D. 76.92 % **

**Description :**

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 %

**9. **
From a homogeneous mixture of salt and sugar containing 4 parts of salt and 5 parts of sugar, 1/4th part of the mixture is drawn off and replaced with salt. The ratio of salt and sugar in the new mixture is?

A. |

B. |

C. |

D. |

**Correct Answer : B. 7 ∶ 5 **

**Description :**

Let amount of the mixture of salt and sugar is 1 kg

Mixture contains 4 parts of salt and 5 parts of sugar

⇒ Amount of salt in the mixture = 4/9 kg

⇒ Amount of sugar in the mixture = 5/9 kg

After replacing 1/4th part with salt,

⇒ Amount of salt in the new mixture = (4/9 – 4/9 × ¼ + 1/4) kg = (16 – 4 + 9) /36 kg = 7/12 kg

⇒ Amount of sugar in the new mixture = (5/9 – 5/9 × 1/4) kg = 5/12 kg

∴ Required ratio = 7/12 ∶ 5/12 = 7 ∶ 5

**10. **
According to an instruction a mixture of colour and Turpentine containing half part of each is perfect for painting a wall. A painter is provided a mixture, 3 parts of which are colour and 5 parts are Turpentine. How much of the mixture drawn off and replaced with colour so that the mixture becomes perfect?

A. |

B. |

C. |

D. |

**Correct Answer : D. 1/5th **

**Description :**

Let total amount of the mixture be 1 litre

3 parts of the mixture are colour and 5 parts are Turpentine

⇒ Amount of colour in the mixture = 3/8 litre

⇒ Amount of turpentine in the mixture = 5/8 litre

Let amount of mixture removed by the painter be ‘x’ litre

⇒ Amount of colour in the new mixture = (3/8 – 3x/8 + x) litre

⇒ Amount of turpentine in the new mixture = (5/8 – 5x/8) litre

The mixture will be perfect if, Amount of colour in the new mixture = Amount of turpentine in the new mixture

⇒ 3/8 – 3x/8 + x = 5/8 – 5x/8

⇒ 3 – 3x + 8x = 5 – 5x

⇒ 10x = 2

⇒ x = 1/5

∴ 1/5th part of the mixture should be replaced