## Alligation or Mixture

#### Mixture :-

When any two or more ingredients are mixed together in a certain ratio , then this new obtained product is called mixture .
Or
Mixture is said to be the combination of two or more quantities .

#### Mean Price :-

The mean or average price of a mixture when the prices of two or more ingredients which may be mixed together and the proportion in which they are mixed are given. Here, cost price of a unit quantity of a mixture is called the ‘mean price’.It will always be higher than cost price of cheaper quantity and lower than cost price of dearer quantity .

#### Alligation :-

Let d per unit be the price of first ingredient (dearer quality) and c per unit be the price of second ingredient (cheaper quality) are mixed with together to form a mixture whose mean price is m per unit, then the two ingredients must be mixed in the ratio :

 Quantity of cheaper = ( Cost Price of dearer - Mean price ) Quantity of dearer ( Mean price − Cost Price of cheaper )
i.e., the two ingredients are to be mixed in the inverse ratio of the differences of their prices and the mean price.
The above rule can be represented by figure given below.

 Quantity of cheaper = ( d - m ) Quantity of dearer ( m - c )

Explanation :- Suppose, p Kg of cheaper quality is mixed with q Kg of dearer quality.
Price of cheaper ingredient = ₹ pc
Price of dearer ingredient = ₹ qd
∴ Price of mixture = ₹ ( pc + qd )
and Total quantity of mixture = ( p + q ) Kg

 ∴ Price of mixture per Kg = ₹ ( pc + qd ) ( p + q )

 ⇒ m = ( pc + qd ) ( p + q )

⇒ ( pc + qd ) = ( mp + mq ) ⇒ ( mp - pc ) = ( qd - mq ) ⇒ p( m - c ) = q( d - m )

 ⇒ p = ( d - m ) q ( m - c )

 Hence , Required ratio = ( d - m ) ( m - c )

Example 1. The cost of type 1 tea is Rs. 72 per kg and type 2 tea is Rs. 62 per kg. Find the ratio in which both are mixed to obtain the mixture of worth Rs. 64.50 per kg? How much should be the quantity of second type of tea, if the first type is 60 Kg.
Sol :- Given , The cost of type 1 tea = Rs. 72 per kg
The cost of type 2 tea = Rs. 62 per kg
Mean cost price of both types of tea = Rs. 64.50 per kg
and
Using Alligation rule ,

∴ Required ratio = 750 : 250 = 3 : 1

 ⇒ Quantity of first type of tea = 3 Quantity of second type of tea 1

 ⇒ 60 = 3 Quantity of second type of tea 1

 ⇒ Quantity of second type of tea = 60 = 20 Kg. 3

Hence , Quantity of second type of tea is 20 Kg.

Example 2. Sugar at ₹ 15 per Kg is mixed with sugar at ₹ 20 per Kg in the ratio 3 : 7. Find the per Kg price of the mixture.
Sol :- Let , the mean price ( m ) of the mixture be ₹ x.
Given , Quantity of cheaper = ₹ 15 per Kg and Quantity of cheaper = ₹ 20 per Kg
Ratio = 3 : 7

Using alligation rule , we have

 ⇒ Quantity of cheaper sugar = 3 Quantity of dearer sugar 7

 ⇒ ( 20 - x ) = 3 ( x - 15 ) 7

⇒ 7( 20 - x ) = 3( x - 15 )
⇒ 140 - 7x = 3x - 45
⇒ 3x + 7x = 140 + 45
⇒ 10x = 185

 ⇒ x = 185 = 18.5 10

Hence , the mean price of the mixture be ₹ 18.5 .

Example 3. To obtain a mixture of Rs. 102/3 a litre, how much water is to be added to 60 litres of milk. The milk is available at 11/2 litre for Rs. 20 .

 Sol :- Given , Mean price of mixture = Rs. 10 2 = Rs. 32 3 3

 Cost price of 1 1 litre milk = Rs. 20 2

 ∴ Cost price of 1 litre milk = Rs. 20 x 2 = Rs. 40 3 3

Using the alligation rule ,

 ∴ Ratio of water and milk = 8 : 32 3 3

 ∴ Quantity of water to be added to 60 litres of milk = 1 × 60 = 15 litres 4

Example 4. The rate per litre of milk is Rs. 12. Find the ratio in which water and milk be mixed to get a mixture of worth Rs. 8 per litre.
Sol :- Given , Mean price = Rs. 8
Cost price of 1 litre of milk = Rs. 12
Let Cost price of 1 litre of water = Rs. 0
Using the alligation rule ,

∴ Ratio of water and milk = 4 : 8 = 1 : 2

#### Some Useful Shortcut Method :-

1. If a container or vessel initially , full of mixture, contains a units of liquid of which b units are withdrawn. The vessel is then filled with b units of another liquid ( water ).Next, b units of the mixture are withdrawn and, again the vessel is filled with b units of liquid. This process is repeated n times.
Then ,

 The final quantity of the original liquid in the vessel or container = a 1 - b n units . a

Or

 Mixture left in the vessel after nth operation = ( a - b ) n units . Original quantity of mixture in the vessel a

Example 5. 4 litres are drawn from a container full of milk and then is filled with water .This operation is performed three more times .The ratio of the quantity of milk left in the container and that of water is 16 : 65 .How much milk did the container hold initially ?
Sol :- Here , b = 4 , n = 4 times and a = ?
Given , Ratio of milk and water = 16 : 65
Using the above given formula ,

 Quantity of milk left in the container after four operations = a 1 - 4 4 units . a

 And Quantity of water left in the container after four operations = a - a 1 - 4 4 units . a

 = a 1 - 1 - 4 4 units . a

According to question ,

 ⇒ [ a( 1 - 4 / a )4 ] = 16 a[ 1 - ( 1 - 4 / a )4 ] 65

 ⇒ [ ( 1 - 4 / a )4 ] = 16 [ 1 - ( 1 - 4 / a )4 ] 65

 ⇒ 1 - 4 4 = 16 1 - 1 - 4 4 a 65 a

 ⇒ 1 - 4 4 = 16 - 16 1 - 4 4 a 65 65 a

 ⇒ 1 - 4 4 + 16 1 - 4 4 = 16 a 65 a 65

 ⇒ 16 + 65 1 - 4 4 = 16 65 a 65

 ⇒ 81 1 - 4 4 = 16 65 a 65

 ⇒ 1 - 4 4 = 16 a 81

 ⇒ 1 - 4 4 = 2 4 a 3

 ⇒ 1 - 4 = 2 a 3

 ⇒ 1 - 2 = 4 3 a

 ⇒ 1 = 4 ⇒ a = 4 x 3 = 12 litres 3 a

2. There are two vessels of equal size filled with mixtures of liquids A and B in the ratio a1 : b1 and a2 : b2 respectively. If the contents of all the vessels are poured into a single large vessel, then

Quantity of liquid A : Quantity of liquid B
 = a1 + a2 : b1 + b2 a1 + b1 a2 + b2 a1 + b1 a2 + b2

Example 6. Two equal cups are filled with mixture of milk and water. The proportion of milk and water in each cup is as follows: In the first cup as 1 : 2 and in the second cup as 3 : 2 .The contents of the two cups are emptied into a single vessel. What is the proportion of milk and water in it?
Sol :- Here , a1 = 1 , a2 = 3 , b1 = 2 and b2 = 2
Using the above given formula ,

∴ Quantity of milk : Quantity of water
 = 1 + 3 : 2 + 2 ( 1 + 2 ) ( 3 + 2 ) ( 1 + 2 ) ( 3 + 2 )

 Quantity of milk : Quantity of water = 1 + 3 : 2 + 2 3 5 3 5

 Quantity of milk : Quantity of water = ( 5 + 9 ) : ( 10 + 6 ) 15 15

Hence , Required ratio of milk and water is 7 : 8.

3. There are n vessels of equal size filled with mixtures of liquids A and B in the ratio a1 : b1 , a2 : b2 ……… , an : bn, respectively. If the contents of all the vessels are poured into a single large vessel, then

Quantity of liquid A : Quantity of liquid B
 = a1 + a2 + ... + an : b1 + b2 + ... + bn a1 + b1 a2 + b2 an + bn a1 + b1 a2 + b2 an + bn

Explanation :- Let the capacity of each vessel be c litres .

 Amount of liquid A in different vessels = a1c , a2c , ... , anc a1 + b1 a2 + b2 an + bn

 Amount of liquid B in different vessels = b1c , b2c , ... , bnc a1 + b1 a2 + b2 an + bn

Now , we have resulting mixture ,

 Amount of liquid A = a1 + a2 + ... + an × c ……. ( 1 ) a1 + b1 a2 + b2 an + bn

 and Amount of liquid B = b1 + b2 + ... + bn × c ……. ( 2 ) a1 + b1 a2 + b2 an + bn

Eq. ( 1 ) divided by eq. ( 2 ) , we get

Quantity of liquid A : Quantity of liquid B
 = a1 + a2 + ... + an × c : b1 + b2 + ... + bn × c a1 + b1 a2 + b2 an + bn a1 + b1 a2 + b2 an + bn

∴ Quantity of liquid A : Quantity of liquid B
 = a1 + a2 + ... + an : b1 + b2 + ... + bn a1 + b1 a2 + b2 an + bn a1 + b1 a2 + b2 an + bn

Example 7. Three equal jars are filled with mixture of milk and water. The proportion of milk and water in each jar is as follows: In the first jar as 1 : 2, in the second jar as 3 : 2 and in the third as 4 : 5. The contents of the three jars are emptied into a single vessel. What is the proportion of milk and water in it?
Sol :- Here , a1 = 1 , a2 = 3 , a3 = 4 , b1 = 2 , b2 = 2 and b3 = 5
Using the above given formula ,
Quantity of milk : Quantity of water

 = 1 + 3 + 4 : 2 + 2 + 5 ( 1 + 2 ) ( 3 + 2 ) ( 4 + 5 ) ( 1 + 2 ) ( 3 + 2 ) ( 4 + 5 )

 Quantity of milk : Quantity of water = 1 + 3 + 4 : 2 + 2 + 5 3 5 9 3 5 9

 Quantity of milk : Quantity of water = 15 + 27 + 20 : 30 + 18 + 25 45 45

∴ Quantity of milk : Quantity of water = 62 : 73.
Hence , Required ratio of milk and water is 62 : 73 .

4. There are n vessels of sizes c1 , c2 ,c3 ……….. , cn filled with mixtures of liquids A and B in the ratio a1 : b1 , a2 : b2 ……… , an : bn, respectively. If the contents of all the vessels are poured into a single large vessel, then

Quantity of liquid A : Quantity of liquid B
 = a1c1 + a2c2 + ... + ancn : b1c1 + b2c2 + ... + bncn a1 + b1 a2 + b2 an + bn a1 + b1 a2 + b2 an + bn

Example 8. Three glasses of sizes 2 litres, 3 litres and 5 litres, contain mixture of milk and water in the ratio 1 : 2, 3 : 4 and 5 : 7 respectively. The contents of all the three glasses are poured into a single vessel. Find out ratio of milk to water in the resulting mixture.
Sol :- Here , c1 = 2 litres , c2 = 3 litres , c3 = 5 litres , a1 = 1 , a2 = 3 , a3 = 5 , b1 = 2 , b2 = 4 and b3 = 7
Using the above given formula ,
Quantity of milk : Quantity of water

 = ( 1 x 2 ) + ( 3 x 3 ) + ( 5 x 5 ) : ( 2 x 2 ) + ( 4 x 3 ) + ( 7 x 5 ) ( 1 + 2 ) ( 3 + 4 ) ( 5 + 7 ) ( 1 + 2 ) ( 3 + 4 ) ( 5 + 7 )

 Quantity of milk : Quantity of water = 56 + 108 + 175 : 112 + 144 + 245 84 84

 Quantity of milk : Quantity of water = 339 : 501 84 84

Hence , Quantity of milk : Quantity of water = 339 : 501

5. In a container , the ratio of milk and water is a : b .If M litre of water is added to this given mixture , the ratio of milk and water becomes a : c . then ,

 Quantity of milk in original mixture = aM litre ( c - b )

 And Quantity of water in original mixture = bM litre ( c - b )

Example 9. In a container , milk and water are present in the ratio 9 : 5. If 10 litre water is added to this mixture , the ratio of milk and water becomes 9 : 7 .Find the quantity of water and milk in new mixture ?
Sol :- Here , a = 9 , b = 5 , c = 7 and m = 10 litre
Using the above given formula , we have

 Quantity of water in original mixture = ( 5 x 10 ) litre = 50 = 25 litre ( 7 - 5 ) 2

Quantity of water in new mixture = 25 + 10 = 35 litre

 and Quantity of milk in original mixture = ( 9 x 10 ) litre = 90 = 45 litre ( 7 - 5 ) 2

Quantity of milk in new mixture = 45 + 10 = 55 litre

Example 10. How many kilograms of a tea worth ₹ 25 per kg must be blended with 30 kg of tea worth ₹ 30 per kg , so that by selling the blended variety at ₹ 30 per kg , there should be a gain of 10% ?
Sol :- Let the quantity of tea worth ₹ 25 = P kg
According to question ,
( 25P + 30 x 30 ) x 110% = 30 ( 30 + P )

 ⇒ ( 25P + 900 ) x 110 = 900 + 30P 100

⇒ ( 275P + 9900 ) = ( 9000 + 300P ) ⇒ 300P - 275P = 9900 - 9000
⇒ 25P = 900

 ⇒ P = 900 = 36 Kg 25

Hence , the quantity of tea worth ₹ 25 be 36 kg .

Example 11. In a mixture of 60 L the ratio of acid and water is 2 : 1. If the ratio of acid and water is to be 1 : 2 , then the amount of water ( in litres ) to be added to the mixture is
Sol :- Given , Total quantity of mixture = 60 L
Ratio of acid and water = 2 : 1

 Quantity of acid in the mixture = 60 × 2 = 40 L 3

 Quantity of water in the mixture = 60 × 1 = 20 L 3

Let P litre quantity of water is to be added , then

 40 = 1 ⇒ 80 = 20 + P ( 20 + P ) 2

⇒ P = 80 - 20 = 60 L
Hence , required quantity of water is 60 L .

Example 12. Tea worth ₹ 126 per kg and ₹ 135 per kg are mixed with a third party variety in the ratio 1 : 1 : 2.If the mixture is worth ₹ 153 per kg, the price of the third variety per kg will be
Sol :- Given , mixed with a third party ratio = 1 : 1 : 2
Let the cost of third variety tea will be ₹ P per kg .
According to question ,

 ( 126 + 135 + 2P ) = 153 4

⇒ 261 + 2P = 4 x 153 ⇒ 261 + 2P = 612 ⇒ 2P = 612 - 261 = 351

 ⇒ P = 351 = 175.50 2

Hence , the cost of third variety tea will be ₹ 175.50 per kg .