###
**Alligation**** and**

Mixtures Important formulas:

Mixtures Important formulas:

**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 a desired

price.

Or, Alligation is an old and practical method of

solving arithmetic problems related to mixtures of ingredients.

solving arithmetic problems related to mixtures of ingredients.

**2. Mean Price:**The cost price of a unit quantity of the mixture is called the mean

price.

**3. Rule of**

Alligation:If two ingredients are mixed, i.e. if

Alligation:

*‘c’*kg of cheaper thing

mixed with ‘

*d*’ kg of dearer thing and we get ‘m’ kg of mixture, then

The Preparation for the solution is like this.

Therefore, (Cheaper quantity): (Dearer quantity) = (d – m): (m – c).

**4.**Suppose a container contains ‘

*x*’ units of liquid from which ‘

*y*’

units are taken out and replaced by water. After ‘n’ operations the quantity of

pure liquid=[

*x*{1- (

*y/x*)}n] units.

__Solved__

Examples:Examples:

__Example:__

0101

**How much water**

must be added to 60 litres of milk at 1½ liter for Rs. 20, so as to have a

mixture worth Rs.10

must be added to 60 litres of milk at 1½ liter for Rs. 20, so as to have a

mixture worth Rs.10

**⅔**

**a liter?**

__Solution:__
C.P. of 1 liter of milk = Rs. (20 × 2/3) = Rs. 40/3

Therefore, Ratio of water and milk = (8/3): (32/3) = 1:

4

4

Hence, Quantity of water to be added

to 60 litres of milk = [(1/4) × 60] Liters

to 60 litres of milk = [(1/4) × 60] Liters

= 15 Liters.

__Example: 02__**How many kgs of wheat**

costing Rs. 8 per kg must be mixed with 86 kg of rice costing Rs. 6.40 per kg

so that 20% gain may be obtained by selling the mixture at Rs. 7.20 per kg ?

costing Rs. 8 per kg must be mixed with 86 kg of rice costing Rs. 6.40 per kg

so that 20% gain may be obtained by selling the mixture at Rs. 7.20 per kg ?

__Solution:__
S.P. of 1 kg mixture = Rs. 7.20, Gain = 20%.

Therefore, C.P. of 1 kg mixture = Rs. [(100/120) × 7.20]

= Rs. 6.

= Rs. 6.

By the rule of alligation, we have:

Wheat of 1st kind: Wheat of 2nd kind = 60: 200 = 3:

10.

10.

Let ‘x’ kg of wheat of 1st kind be mixed with 36 kg of

wheat of 2nd kind.

wheat of 2nd kind.

Then, 3: 10 =

*x*: 36 or 10*x*= 3 × 36 or*x*= 10.8 kg.**SOLVED EXAMPLES**

**PART 1**

__Question__

No. 01No. 01

**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

containing half milk and half water?

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

containing half milk and half water?

__Solution:__
Let the C.P. of milk be Rs. 1 per litre

Milk in 1 litre mixture of A = 4/7 litre; Milk in 1

litre mixture of B = 2/5 litre;

litre mixture of B = 2/5 litre;

Milk in 1 litre mixture of C = ½ litre

C.P. of 1 litre mixture in A = Rs. 4/7

C.P. of

1 litre mixture in B = Rs. 2/5

1 litre mixture in B = Rs. 2/5

Mean price = Rs. 1/2

By the rule of alligation, we have:

Allegation and Mixture Solution-01 |

Therefore, required ratio = (1/10): (1/14) = 7: 5

__Question__

No. 02No. 02

**In what ratio must water be**

mixed with milk to gain 20 % by selling the mixture at cost price?

mixed with milk to gain 20 % by selling the mixture at cost price?

__Solution:__
Let C.P. of milk be Rs. 1 per litre.

Then, S.P. of 1 litre of mixture = Rs. 1.

Gain obtained = 20%.

Therefore, C.P. of 1 litre of mixture = Rs. [(100/120)

× 1] = 5/6

× 1] = 5/6

By the rule of alligation, we have:

Alligation and Mixture Solution-02 |

Therefore, Ratio of water and milk = (1/6): (5/6) = 1:

5.

5.

__Question__

No. 03No. 03

**In what ratio**

must rice at Rs. 9.30 per kg be mixed with rice at Rs. 10.80 per kg so that the

mixture be worth Rs. 10 per kg?

must rice at Rs. 9.30 per kg be mixed with rice at Rs. 10.80 per kg so that the

mixture be worth Rs. 10 per kg?

__Sol__**By the rule of alligation, we have:**

__ution:__Alligation and Mixture Solution-03 |

(Cheaper quantity): (Dearer quantity) =

(d – m): (m – c).

(d – m): (m – c).

=> (1080 – 1000): (1000 – 930) => 80: 70 =>8: 7

Therefore, required ratio =8: 7.

__Question__

No. 04No. 04

**A vessel is filled with liquid, 3 parts**

of which are water and 5 parts acid. How much of the mixture must be drawn off

and replaced with water so that the mixture may be half water and half acid?

of which are water and 5 parts acid. How much of the mixture must be drawn off

and replaced with water so that the mixture may be half water and half acid?

__Solution:__
Suppose the vessel initially contains 8

litres of liquid.

litres of liquid.

Let, ‘x’ litres of this liquid be

replaced with water.

replaced with water.

Quantity of water in new mixture = [3 –

(3

(3

*x*/8) +*x*] litres
Quantity of acid in new mixture = [5 –

(5

(5

*x*/8)] litres
Therefore, [3 – (3

(5

*x*/8) +*x*] = [5 –(5

*x*/8)]
=> 5

=>

*x*+ 24 = 40 – 5*x*=> 10*x*= 16=>

*x*= 8/5
So, parts of mixture replaced = (8/5) ×

(1/8) = 1/5.

(1/8) = 1/5.

__Question__

No. 05No. 05

**Coffee**

wroth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the

ratio 1: 1: 2. If the mixture is worth Rs. 153 per kg, the price of the third

variety per kg will be:

wroth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the

ratio 1: 1: 2. If the mixture is worth Rs. 153 per kg, the price of the third

variety per kg will be:

__Solution:__
Since 1st and 2nd variety is mixed in

equal proportions.

equal proportions.

So, their average price = Rs. [126 +

135)/2] = Rs. 130.50

135)/2] = Rs. 130.50

So, the mixture is formed by mixing two verities,

one at Rs. 135.50 per kg and the other at say, Rs.

2, i.e. the price of 3rd kind is Rs.

one at Rs. 135.50 per kg and the other at say, Rs.

*x*per kg in the ratio of 2 :2, i.e. the price of 3rd kind is Rs.

*x*.
By the rule of alligation, we have:

Al ligation and Mixture Solution-05 |

Therefore, (

175.50

*x*– 153) = 22.50 =>*x*=175.50

Hence, the price of 3rd variety per kg is

Rs. 175.50

Rs. 175.50

__Question__

No. 06No. 06

**In**

what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20

per kg respectively so as to get a mixture worth Rs. 16.50 kg?

what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20

per kg respectively so as to get a mixture worth Rs. 16.50 kg?

__Solution:__
By the rule of alligation, we have:

Alligation and Mixture Solution-06 |

Therefore, the required rate = 3.50: 1.50

= 7: 3

= 7: 3

__Question__

No. 07No. 07

A merchant has 1000 kg of sugar, part of

which he sells at 8% profit and the rest at 18% profit. He gains 14% on the

whole. The quantity sold at 18% profit is:

which he sells at 8% profit and the rest at 18% profit. He gains 14% on the

whole. The quantity sold at 18% profit is:

__Solution:__
By the rule of alligation, we have:

Alligation and Mixture Solution-07 |

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

= 2: 3

= 2: 3

Therefore, the quantity of 2nd kind =

(3/5) × 1000 = 600 kg.

(3/5) × 1000 = 600 kg.

__Question__

No. 08No. 08

**A dishonest professes to sell his milk at**

cost price but he mixes it with water and thereby gains 25%. The percentage of

water in the mixture is:

cost price but he mixes it with water and thereby gains 25%. The percentage of

water in the mixture is:

__Solution:__
Let, C.P of 1 litre milk be Rs. 1

Then, S.P of 1 litre of mixture = Rs. 1,

Gain = 25%

Gain = 25%

Therefore, C.P of 1 litre mixture = Rs.

(100/125) × 1 = 4/5

(100/125) × 1 = 4/5

By the rule of alligation, we have:

Alligation and Mixture Solution-08 |

Therefore, ratio of milk to water = (4/5):

(1/5)

(1/5)

Hence, percentage of water in the mixture

= [(1/5) × 100] % = 20%

= [(1/5) × 100] % = 20%

__Question__

No. 09No. 09

**A jar full of water contains 40% sugar**

dissolved in it. A part of the water is replaced by another containing 19% of

sugar and now the percentage of sugar dissolved was found to be 26%. The

quantity of sugar replaced is:

dissolved in it. A part of the water is replaced by another containing 19% of

sugar and now the percentage of sugar dissolved was found to be 26%. The

quantity of sugar replaced is:

__Solution:__
By the rule of alligation, we have:

Alligation and Mixture Solution-09 |

So, the ratio of 1st and 2nd quantities =

7: 14 = 1: 2

7: 14 = 1: 2

Therefore, required quantity replaced =

2/3

2/3

__Question__

No. 10No. 10

**The cost of type 1 rice is Rs. 15 per kg**

and type 2 rice is Rs. 20 per kg. If both type 1 and type 2 are mixed in the

ratio of 2: 3, then the price per kg of the mixed variety of rice is:

and type 2 rice is Rs. 20 per kg. If both type 1 and type 2 are mixed in the

ratio of 2: 3, then the price per kg of the mixed variety of rice is:

__Solution:__
Let the price of the mixed variety be Rs.

*x*per kg.
By the rule of alligation, we have:

Alligation and Mixture Solution-10 |

Therefore, (20 –

*x*)/ (*x*– 15) = 2/3
=> (60 – 3

*x*) = (2*x*– 30)
=> 5

*x*= 90
=>

*x*= 18
Therefore, the price of the mixed variety

is Rs. 18 per kg.

is Rs. 18 per kg.

**PART 2**

__Question No. 01__**A cup**

contains milk and water in the ratio 3:1. How much mixture should be taken out

and water added to make the ratio 1:1?

contains milk and water in the ratio 3:1. How much mixture should be taken out

and water added to make the ratio 1:1?

(A) 1/3 part

of mixture

of mixture

(B) 1/2 part

of mixture

of mixture

(C) 1/4 part

of mixture

of mixture

(D) 1/5 part

of mixture

of mixture

Answer:

Option A

Option A

__Explanation:__
Let

*y*part of mixture is replaced by water
Therefore,

(4

(4

*x*–*y*)/4*x*= 2*x*/3*x*
=> 12

*x*² – 3*xy*= 8*x*²
=>

*y*= 4*x*/3
I.e. 1/3

part of mixture.

part of mixture.

__Question No. 02__**A milk**

vendor has 2 cans of milk. The first contains 25% water and the rest milk. The

second contains 50% water. How much milk should he mix from each of the

containers so as to get 12 liters of milk such that the ratio of water to milk

is 3:5?

vendor has 2 cans of milk. The first contains 25% water and the rest milk. The

second contains 50% water. How much milk should he mix from each of the

containers so as to get 12 liters of milk such that the ratio of water to milk

is 3:5?

(A) 4

liters, 8 liters

liters, 8 liters

(B) 6

liters, 6 liters

liters, 6 liters

(C) 5

liters, 7 liters

liters, 7 liters

(D) 7

liters, 5 liters

liters, 5 liters

Answer:

Option B

Option B

__Explanation:__
Let the cost

of 1 liter milk be Rs. 1

of 1 liter milk be Rs. 1

Milk in 1

liter mix in 1st can = 3/4 liter, C.P of 1 liter mix in 1st can Rs. 3/4

liter mix in 1st can = 3/4 liter, C.P of 1 liter mix in 1st can Rs. 3/4

Milk in 1

liter mix in 2nd can = 1/2 liter, C.P of 1 liter mix in 2nd can Rs. 1/2

liter mix in 2nd can = 1/2 liter, C.P of 1 liter mix in 2nd can Rs. 1/2

Milk in 1

liter of final mix = 5/8 liter, Mean price = Rs. 5/8

liter of final mix = 5/8 liter, Mean price = Rs. 5/8

By the rule

of alligation, we have:

∴ Ratio

of two mixtures = (1/8) : (1/8) = 1 : 1

of two mixtures = (1/8) : (1/8) = 1 : 1

So, quantity

of mixture taken from each can = (½ × 12) = 6 liters

of mixture taken from each can = (½ × 12) = 6 liters

__Question No. 03__**In 2 gallons**

mixture of spirit and water, the percentage of water is 12. In another 3

gallons mixture of spirit and water, the percentage of water is 5. These two

mixtures are poured in a third pot and ½ gallon more water is added to it. Find

the percentage of water in this new mixture.

mixture of spirit and water, the percentage of water is 12. In another 3

gallons mixture of spirit and water, the percentage of water is 5. These two

mixtures are poured in a third pot and ½ gallon more water is added to it. Find

the percentage of water in this new mixture.

(A) 17 1

*/*11 %
(B) 17 3

*/*11 %
(C) 16 2

*/*11 %
(D) 16 13

*/*27 %
Answer:

Option C

Option C

__Explanation:__
W1 =

12% of 2 = 24/100 = 0.24

12% of 2 = 24/100 = 0.24

∴ W2

= 5% of 3 = 0.15

= 5% of 3 = 0.15

W3 = 0.5

∴ W1

+ W2 + W3 = 0.24 + 0.15 + 0.5 = 0.89

+ W2 + W3 = 0.24 + 0.15 + 0.5 = 0.89

∴ New

W% = (0.89/5.5) × 100 = 16 2

%

W% = (0.89/5.5) × 100 = 16 2

*/*11%

__Question No. 04__**How many kg**

of sugar costing Rs. 9 per kg. must be mixed with 27 kg. of sugar costing Rs. 7

per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24

per kg.?

of sugar costing Rs. 9 per kg. must be mixed with 27 kg. of sugar costing Rs. 7

per kg so that there may be a gain of 10% by selling the mixture at Rs. 9.24

per kg.?

(A) 36 kg.

(B) 42 kg.

(C) 54 kg.

(D) 63 kg.

Answer:

Option D

Option D

__Explanation:__
S.P of 1 kg

of mixture = Rs. 9.24, Gain 10%

of mixture = Rs. 9.24, Gain 10%

∴ C.P

of 1 kg of mixture = Rs. [(100/110) × 9.24] = Rs. 8.40

of 1 kg of mixture = Rs. [(100/110) × 9.24] = Rs. 8.40

By the rule

of alligation, we have:

of alligation, we have:

∴ Ratio

of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3

of quantities of 1st and 2nd kind = 14 : 6 = 7 : 3

Let,

27 kg of 2nd kind

*x*kg of sugar of 1st kind be mixed with27 kg of 2nd kind

Then, 7 : 3

=

=

*x*: 27
=>

*x*= (7 × 27)/3 = 63 kg.

__Question No. 05__**Three**

identical vessels contain the mixture of spirit and water. The ratio of spirit

and water in each glass is 2:3, 3:4 and 4:5 respectively. The mixtures of all

the three vessels are poured into a big pot. The ratio of spirit and water in

the new mixture is

identical vessels contain the mixture of spirit and water. The ratio of spirit

and water in each glass is 2:3, 3:4 and 4:5 respectively. The mixtures of all

the three vessels are poured into a big pot. The ratio of spirit and water in

the new mixture is

(A) 401/544

(B) 27/37

(C) 19/37

(D) 13/37

Answer:

Option A

Option A

__Explanation:__
Net spirit =

(2/5) + (3/7) + (4/9) = 401/315

(2/5) + (3/7) + (4/9) = 401/315

∴ Net

water = 3 – (401/315) = 544/325

water = 3 – (401/315) = 544/325

Hence

required ratio = 401/544

required ratio = 401/544

__Question No. 06__**A container**

contains 40 liters of milk. From this container 4 liters of milk was taken out

and replaced with water. This process was repeated further 2 times. How much

milk is now contained by the container?

contains 40 liters of milk. From this container 4 liters of milk was taken out

and replaced with water. This process was repeated further 2 times. How much

milk is now contained by the container?

(A) 26.34

liters

liters

(B) 27.36

liters

liters

(C) 28

liters

liters

(D) 29.16

liters

liters

Answer:

Option D

Option D

__Explanation:__
Amount of

milk left after 3 operations = [40 × {1 – (4/40)}3] liters

milk left after 3 operations = [40 × {1 – (4/40)}3] liters

= [40 × (9/10) × (9/10) × (9/10)] = 29.16 liters

__Question No. 07__**In what**

ratio must water be mixed with milk to gain 16⅔ % on selling the mixture at

cost price?

ratio must water be mixed with milk to gain 16⅔ % on selling the mixture at

cost price?

(A) 1 : 6

(B) 6 : 1

(C) 2 : 3

(D) 4 : 3

Answer:

Option A

Option A

__Explanation:__
Let C.P of 1

liter milk be Rs.1

liter milk be Rs.1

S.P of 1

liter of mixture = Rs. 1, gain = (50/3) %

liter of mixture = Rs. 1, gain = (50/3) %

∴ C.P

of 1 liter of mixture = [100 × (3/350) × 1] = 6/7

of 1 liter of mixture = [100 × (3/350) × 1] = 6/7

By the rule

of alligation, we have:

of alligation, we have:

∴ Ratio

of water and milk = (1/7) : (6/7) = 1 : 6

of water and milk = (1/7) : (6/7) = 1 : 6

__Question No. 08__**Find the**

ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to

produce a mixture worth Rs. 6.30 a kg.

ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to

produce a mixture worth Rs. 6.30 a kg.

(A) 1 : 3

(B) 2 : 3

(C) 3 : 4

(D) 4 : 5

Answer:

Option B

Option B

__Explanation:__
By the rule

of alligation, we have:

of alligation, we have:

∴ Required

ratio = 60 : 90 = 2 : 3

ratio = 60 : 90 = 2 : 3

__Question No. 09__**In what**

ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg

so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg

so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

(A) 3 : 2

(B) 3 : 4

(C) 3 : 5

(D) 4 : 5

Answer:

Option A

Option A

__Explanation:__
S.P of 1 kg

of mixture = 68.20, Gain = 10%

of mixture = 68.20, Gain = 10%

C.P of 1 kg

of the mixture = Rs [(100/110) × 68.20] = Rs. 62

of the mixture = Rs [(100/110) × 68.20] = Rs. 62

By the rule

of alligation, we have:

of alligation, we have:

∴ Required

ratio = 3 : 2

ratio = 3 : 2

__Question No. 10__**A can**

contains a mixture of two liquids ‘A’ and ‘B’ in the ratio 7 : 5. When 9 liters

of mixture are drawn off and the can is filled with ‘B’, the ratio of ‘A’ and

‘B’ becomes 7 : 9. how many liters of liquid ‘A’ was contained by the can

initially?

contains a mixture of two liquids ‘A’ and ‘B’ in the ratio 7 : 5. When 9 liters

of mixture are drawn off and the can is filled with ‘B’, the ratio of ‘A’ and

‘B’ becomes 7 : 9. how many liters of liquid ‘A’ was contained by the can

initially?

(A) 10

(B) 20

(C) 21

(D) 25

Answer:

Option C

Option C

__Explanation:__
Suppose the

can initially contains 7

can initially contains 7

*x*and 5*x*of mixtures*A*and*B*respectively
Quantity of

A in mixture left = [7

× 9)] liters = [7

A in mixture left = [7

*x*– (7/12× 9)] liters = [7

*x*– (21/4)] liters
Quantity of

B in the mixture left = [5

× 9)] liters = [5

B in the mixture left = [5

*x*– (5/12× 9)] liters = [5

*x*– (15/4)] liters
∴ [7

*x*– (21/4)]/ [5*x*– (15/4)] = 7/9
=> (28

*x*– 21)/(20*x*+ 21) = 7/9
=> 252

*x*– 189 = 140*x*+ 147
=> 112

*x*= 336
=>

*x*= 3
So, the can

contained 21 liters of A.

contained 21 liters of A.

__Question No. 11__**8 liters are**

drawn from a cask full of wine and is then filled with water. This operation is

performed three more times. The ratio of the quantity of wine now left in cask

to that of water is 16:65. How much wine did the cask hold originally?

drawn from a cask full of wine and is then filled with water. This operation is

performed three more times. The ratio of the quantity of wine now left in cask

to that of water is 16:65. How much wine did the cask hold originally?

(A) 18

liters

liters

(B) 24

liters

liters

(C) 32

liters

liters

(D) 42

liters

liters

Answer:

Option B

Option B

__Explanation:__
Let the

quantity of the wine in the cask originally be

quantity of the wine in the cask originally be

*x*liters
Then,

quantity of wine left in the cask after 4 operations = [

liters

quantity of wine left in the cask after 4 operations = [

*x*{1 – (8/*x*)}4]liters

[

16/81

*x*{1 – (8/*x*)}4]/*x*=16/81

=> {1 –

(8/

(8/

*x*)}4 = (2/3)4
=> [(

*x*– 8)/*x*] = 2/3
=> 3

*x*– 24 = 2*x*
=>

*x*= 24