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**Alligation**** and
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.

**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

*‘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:__

__Example: 01__**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**

**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

Hence, Quantity of water to be added
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 ?**

__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.

By the rule of alligation, we have:

Wheat of 1st kind: Wheat of 2nd kind = 60: 200 = 3:
10.

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

__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;

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

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. 02__**In what ratio must water be 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

By the rule of alligation, we have:

Alligation and Mixture Solution-02 |

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

__Question No. 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?**

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

__ution:__Alligation and Mixture Solution-03 |

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

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

Therefore, required ratio =8: 7.

__Question No. 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?**

__Solution:__
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 = [3 -
(3

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

*x*/8)] litres
Therefore, [3 - (3

*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.

__Question No. 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:**

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

So, their average price = Rs. [126 +
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.

*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, (

*x*- 153) = 22.50 =>*x*= 175.50
Hence, the price of 3rd variety per kg is
Rs. 175.50

__Question No. 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?**

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

Alligation and Mixture Solution-06 |

Therefore, the required rate = 3.50: 1.50
= 7: 3

__Question No. 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:

__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

Therefore, the quantity of 2nd kind =
(3/5) × 1000 = 600 kg.

__Question No. 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:**

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

Then, S.P of 1 litre of mixture = Rs. 1,
Gain = 25%

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

Hence, percentage of water in the mixture
= [(1/5) × 100] % = 20%

__Question No. 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:**

__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

Therefore, required quantity replaced =
2/3

__Question No. 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:**

__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.

**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?**

(A) 1/3 part
of mixture

(B) 1/2 part
of mixture

(C) 1/4 part
of mixture

(D) 1/5 part
of mixture

Answer:
Option A

__Explanation:__
Let

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

__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?**

(A) 4
liters, 8 liters

(B) 6
liters, 6 liters

(C) 5
liters, 7 liters

(D) 7
liters, 5 liters

Answer:
Option B

__Explanation:__
Let the cost
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

Milk in 1
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

By the rule
of alligation, we have:
∴ Ratio
of two mixtures = (1/8) : (1/8) = 1 : 1

So, quantity
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.**

(A) 17

^{1}*/*_{11}%
(B) 17

^{3}*/*_{11}%
(C) 16

^{2}*/*_{11}%
(D) 16

^{13}*/*_{27}%
Answer:
Option C

__Explanation:__
W

_{1 }= 12% of 2 = 24/100 = 0.24
∴ W

_{2}= 5% of 3 = 0.15
W

_{3}= 0.5
∴ W

_{1}+ W_{2}+ W_{3}= 0.24 + 0.15 + 0.5 = 0.89
∴ New
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.?**

(A) 36 kg.

(B) 42 kg.

(C) 54 kg.

(D) 63 kg.

Answer:
Option D

__Explanation:__
S.P of 1 kg
of mixture = Rs. 9.24, Gain 10%

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

By the rule
of alligation, we have:

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

Let,

*x*kg of sugar of 1st kind be mixed with 27 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**

(A) 401/544

(B) 27/37

(C) 19/37

(D) 13/37

Answer:
Option A

__Explanation:__
Net spirit =
(2/5) + (3/7) + (4/9) = 401/315

∴ Net
water = 3 - (401/315) = 544/325

Hence
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?**

(A) 26.34
liters

(B) 27.36
liters

(C) 28
liters

(D) 29.16
liters

Answer:
Option D

__Explanation:__
Amount of
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?**

(A) 1 : 6

(B) 6 : 1

(C) 2 : 3

(D) 4 : 3

Answer:
Option A

__Explanation:__
Let C.P of 1
liter milk be Rs.1

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

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

By the rule
of alligation, we have:

∴ Ratio
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.**

(A) 1 : 3

(B) 2 : 3

(C) 3 : 4

(D) 4 : 5

Answer:
Option B

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

∴ Required
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%?**

(A) 3 : 2

(B) 3 : 4

(C) 3 : 5

(D) 4 : 5

Answer:
Option A

__Explanation:__
S.P of 1 kg
of mixture = 68.20, Gain = 10%

C.P of 1 kg
of the mixture = Rs [(100/110) × 68.20] = Rs. 62

By the rule
of alligation, we have:

∴ Required
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?**

(A) 10

(B) 20

(C) 21

(D) 25

Answer:
Option C

__Explanation:__
Suppose the
can initially contains 7

*x*and 5*x*of mixtures*A*and*B*respectively
Quantity of
A in mixture left = [7

*x*- (^{7}/_{12}× 9)] liters = [7*x*- (21/4)] liters
Quantity of
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.

__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?**

(A) 18
liters

(B) 24
liters

(C) 32
liters

(D) 42
liters

Answer:
Option B

__Explanation:__
Let the
quantity of the wine in the cask originally be

*x*liters
Then,
quantity of wine left in the cask after 4 operations = [

*x*{1 - (8/*x*)}^{4}] liters
[

*x*{1 - (8/*x*)}^{4}]/*x*= 16/81
=> {1 -
(8/

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

*x*- 8)/*x*] = 2/3
=> 3

*x*- 24 = 2*x*
=>

*x*= 24

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