###### 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

Question 1

Milk and water are mixed in a vessel A as 4:1 and in vessel B as 3:2. For vessel C, if one takes equal quantities from A and B, find the ratio of milk to water in C.

A. |

B. |

C. |

D. |

**Correct Answer : D. 7:3 **

**Description :**

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

=> \( 4 \over 5 \) : \( 3 \over 5 \)

x

\( 3 \over 5 \) - x x - \( 4 \over 5 \)

=> x = \( 7 \over 10 \)

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

Question 2

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 %

Question 3

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?

A. |

B. |

C. |

D. |

**Correct Answer : D. 7:5 **

**Description :**

Milk in 1-litre mixture of A = 4/7 litre.

Milk in 1-litre mixture of B = 2/5 litre.

Milk in 1-litre mixture of C = 1/2 litre.

By rule of alligation we have required ratio X:Y

X : Y 4/7 2/5 \ / (Mean ratio) (1/2) / \ (1/2 – 2/5) : (4/7 – 1/2) 1/10 1/1 4

So Required ratio = X : Y = 1/10 : 1/14 = 7:5

Question 4

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 : B. 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.

Question 5

An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:

A. |

B. |

C. |

D. |

**Correct Answer : A. \( {1 \over 8} \) kg **

**Description :**

Ratio of Zinc, Copper and Tin is given as,
Z : C : T = 2 : 3 : 1.

Now, let the first alloy be 12 kg (taken as 4 kg Zinc, 6 kg Copper and 2 Kg Lead). Weight of second alloy = 12 kg as, C : T : L = 5 : 4 : 3. (taken as 5 kg Copper, 4 kg Tin and 3 Kg Lead.)

Alloys are mixed together to form third alloy. Then the ratio of content in it,

Z : C : T : L = 4 : (6 + 5) : (2 + 4) : 3

Weight of third alloy = 12 + 12 = 24 Kg.

So, weight of the Lead = \( {3 \over 24} \) = \( {1 \over 8} \) kg.

Question 6

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 : D. 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

Question 7

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

Question 8

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 : A. 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.

Question 9

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 : C. 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.

Question 10

The ratio of Iron and Zinc in an alloy is 4 : 5. In another alloy, the ratio of Iron, Copper and Zinc is 3 : 2 : 7. Equal amounts of the two alloys are molten and mixed together. What will be the ratio of Iron, Copper and Zinc in the resultant alloy?

A. |

B. |

C. |

D. |

**Correct Answer : A. 25 : 6 : 41 **

**Description :**

In first alloy, ratio of iron and zinc is 4 : 5

In second alloy, ratio of iron, copper and zinc is 3 : 2 : 7

Suppose, amount T of both alloys is taken.

⇒ Amount of iron in first alloy = (4/ (4 + 5)) × T = (4/9) T

Amount of zinc in first alloy = (5/ (4 + 5)) × T = (5/9) T

Amount of iron in Second alloy = (3/ (3 + 2 + 7)) × T = (1/4) T

Amount of Copper in second alloy = (2/ (3 + 2 + 7)) × T = (1/6) T

Amount of zinc in second alloy = (7/ (3 + 2 + 7)) × T = (7/12) T

After mixing

Total amount of Iron = (4/9) T + (1/4) T = (25/36) T

Total amount of copper = (1/6) T = (6/36) T

Total amount of zinc = (5/9) T + (7/12) T = (41/36) T

⇒ Ratio of Iron, Copper and Zinc = (25/36) T : (6/36) T : (41/36) T

= 25 : 6 : 41